Exercise 1.2 — Class 9 Mathematics Solutions

Solution

1. Identify the conjugate of the denominator.

   The denominator is $4+\sqrt{3}$. Its conjugate is $4-\sqrt{3}$.

2. Multiply numerator and denominator by the conjugate.

   $$\frac{13}{4+\sqrt{3}}\times \frac{4-\sqrt{3}}{4-\sqrt{3}}=\frac{13(4-\sqrt{3})}{(4+\sqrt{3})(4-\sqrt{3})}$$

3. Apply the difference of squares identity $(a+b)(a-b)=a^2-b^2$.

   $$\frac{13(4-\sqrt{3})}{4^2-(\sqrt{3}{)}^2}=\frac{13(4-\sqrt{3})}{16-3}=\frac{13(4-\sqrt{3})}{13}$$

4. Simplify the fraction.

   Canceling the common factor $13$ from the numerator and denominator:

   $$=4-\sqrt{3}$$

Solution

1. Identify the rationalizing factor.

   The monomial denominator is $\sqrt{3}$. We multiply both numerator and denominator by $\sqrt{3}$.

2. Multiply numerator and denominator by $\sqrt{3}$.

   $$\frac{\sqrt{2}+\sqrt{5}}{\sqrt{3}}\times \frac{\sqrt{3}}{\sqrt{3}}=\frac{(\sqrt{2}+\sqrt{5})\sqrt{3}}{\sqrt{3}\times \sqrt{3}}$$

3. Distribute $\sqrt{3}$ across the numerator.

   $$\frac{\sqrt{2\times 3}+\sqrt{5\times 3}}{3}=\frac{\sqrt{6}+\sqrt{15}}{3}$$

Solution

1. Multiply numerator and denominator by $\sqrt{5}$.

   $$\frac{\sqrt{2}-1}{\sqrt{5}}\times \frac{\sqrt{5}}{\sqrt{5}}=\frac{(\sqrt{2}-1)\sqrt{5}}{(\sqrt{5}{)}^2}$$

2. Distribute and simplify.

   $$\frac{\sqrt{2}\times \sqrt{5}-1\times \sqrt{5}}{5}=\frac{\sqrt{10}-\sqrt{5}}{5}$$

Solution

1. Multiply numerator and denominator by conjugate $6-4\sqrt{2}$.

   $$\frac{6-4\sqrt{2}}{6+4\sqrt{2}}\times \frac{6-4\sqrt{2}}{6-4\sqrt{2}}=\frac{(6-4\sqrt{2}{)}^2}{(6+4\sqrt{2})(6-4\sqrt{2})}$$

2. Expand the numerator using $(a-b{)}^2=a^2-2ab+b^2$ and denominator using $a^2-b^2$.

   $$\text{Numerator}=6^2-2(6)(4\sqrt{2})+(4\sqrt{2}{)}^2=36-48\sqrt{2}+16(2)=36-48\sqrt{2}+32=68-48\sqrt{2}$$ $$\text{Denominator}=6^2-(4\sqrt{2}{)}^2=36-16(2)=36-32=4$$

3. Divide each term by the denominator $4$.

   $$\frac{68-48\sqrt{2}}{4}=\frac{68}{4}-\frac{48\sqrt{2}}{4}=17-12\sqrt{2}$$

Solution

1. Multiply numerator and denominator by conjugate $\sqrt{3}-\sqrt{2}$.

   $$\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{(\sqrt{3}-\sqrt{2}{)}^2}{(\sqrt{3}+\sqrt{2})(\sqrt{3}-\sqrt{2})}$$

2. Expand numerator and denominator.

   $$\text{Numerator}=(\sqrt{3}{)}^2-2(\sqrt{3})(\sqrt{2})+(\sqrt{2}{)}^2=3-2\sqrt{6}+2=5-2\sqrt{6}$$ $$\text{Denominator}=(\sqrt{3}{)}^2-(\sqrt{2}{)}^2=3-2=1$$

3. Write the final simplified form.

   $$\frac{5-2\sqrt{6}}{1}=5-2\sqrt{6}$$

Solution

1. Multiply numerator and denominator by conjugate $\sqrt{7}-\sqrt{5}$.

   $$\frac{4\sqrt{3}}{\sqrt{7}+\sqrt{5}}\times \frac{\sqrt{7}-\sqrt{5}}{\sqrt{7}-\sqrt{5}}=\frac{4\sqrt{3}(\sqrt{7}-\sqrt{5})}{(\sqrt{7}{)}^2-(\sqrt{5}{)}^2}$$

2. Simplify the denominator.

   $$(\sqrt{7}{)}^2-(\sqrt{5}{)}^2=7-5=2$$

3. Cancel the common factor $2$.

   $$\frac{4\sqrt{3}(\sqrt{7}-\sqrt{5})}{2}=2\sqrt{3}(\sqrt{7}-\sqrt{5})$$

Solution

1. Convert negative exponent to positive by taking the reciprocal.

   $${\left(\frac{81}{16}\right)}^{-\frac{3}{4}}={\left(\frac{16}{81}\right)}^{\frac{3}{4}}$$

2. Express $16$ and $81$ as powers of $2$ and $3$.

   Since $16=2^4$ and $81=3^4$:

   $$\frac{16}{81}={\left(\frac{2}{3}\right)}^4$$

3. Apply the power of a power rule $(a^m{)}^n=a^{mn}$.

   $${\left[{\left(\frac{2}{3}\right)}^4\right]}^{\frac{3}{4}}={\left(\frac{2}{3}\right)}^{4\times \frac{3}{4}}={\left(\frac{2}{3}\right)}^3$$

4. Calculate the final value.

   $${\left(\frac{2}{3}\right)}^3=\frac{2^3}{3^3}=\frac{8}{27}$$

Solution

1. Simplify each component into prime factors of $2$ and $3$.

   • ${\left(\frac{3}{4}\right)}^{-2}={\left(\frac{4}{3}\right)}^2=\frac{4^2}{3^2}=\frac{(2^2{)}^2}{3^2}=\frac{2^4}{3^2}$
   • ${\left(\frac{4}{9}\right)}^3=\frac{4^3}{9^3}=\frac{(2^2{)}^3}{(3^2{)}^3}=\frac{2^6}{3^6}$
   • $\frac{16}{27}=\frac{2^4}{3^3}$

2. Change division into multiplication by inverting ${\left(\frac{4}{9}\right)}^3$.

   $$\frac{2^4}{3^2}\times \frac{3^6}{2^6}\times \frac{2^4}{3^3}$$

3. Combine powers of like bases using exponent laws.

   $$\frac{2^{4+4}\times 3^6}{2^6\times 3^{2+3}}=\frac{2^8\times 3^6}{2^6\times 3^5}=2^{8-6}\times 3^{6-5}=2^2\times 3^1$$

4. Compute the final product.

   $$2^2\times 3=4\times 3=12$$

Solution

1. Express decimal as a fraction.

   $$0.027=\frac{27}{1000}$$

2. Express $27$ and $1000$ as cubes.

   $$\frac{27}{1000}={\left(\frac{3}{10}\right)}^3$$

3. Apply the power $-\frac{1}{3}$.

   $$(0.027{)}^{-\frac{1}{3}}={\left[{\left(\frac{3}{10}\right)}^3\right]}^{-\frac{1}{3}}={\left(\frac{3}{10}\right)}^{3\times (-\frac{1}{3})}={\left(\frac{3}{10}\right)}^{-1}$$

4. Invert the fraction.

   $${\left(\frac{3}{10}\right)}^{-1}=\frac{10}{3}$$

Solution

1. Simplify the terms inside the radical using $\frac{a^m}{a^n}=a^{m-n}$.

   $$\frac{x^{14}\cdot y^{21}\cdot z^{35}}{y^{14}\cdot z^7}=x^{14}\cdot y^{21-14}\cdot z^{35-7}=x^{14}{y}^7{z}^{28}$$

2. Convert radical to rational exponent $\frac{1}{7}$.

   $$\sqrt[7]{x^{14}{y}^7{z}^{28}}={\left(x^{14}{y}^7{z}^{28}\right)}^{\frac{1}{7}}$$

3. Distribute the exponent to each variable.

   $$x^{14\times \frac{1}{7}}\cdot y^{7\times \frac{1}{7}}\cdot z^{28\times \frac{1}{7}}=x^2{y}^1{z}^4=x^{2}y{z}^4$$

Solution

1. Express all bases in terms of prime base $5$.

   Since $25=5^2$:

   • $(25{)}^{n+1}=(5^2{)}^{n+1}=5^{2n+2}$
   • $25\cdot (5{)}^{2n}=5^2\cdot 5^{2n}=5^{2n+2}$
   • $5\cdot (5{)}^{2n+3}=5^1\cdot 5^{2n+3}=5^{2n+4}$

2. Substitute into the numerator and denominator.

   $$\text{Numerator}=5^1\cdot 5^{2n+2}-5^{2n+2}=5^{2n+3}-5^{2n+2}=5^{2n+2}(5^1-1)=5^{2n+2}(4)$$ $$\text{Denominator}=5^{2n+4}-5^{2n+2}=5^{2n+2}(5^2-1)=5^{2n+2}(25-1)=5^{2n+2}(24)$$

3. Cancel common factors and simplify.

   $$\frac{5^{2n+2}\times 4}{5^{2n+2}\times 24}=\frac{4}{24}=\frac{1}{6}$$

Solution

1. Express all terms with base $2$.

   • $(16{)}^{x+1}=(2^4{)}^{x+1}=2^{4x+4}=2^{4x}\cdot 2^4=16\cdot 2^{4x}$
   • $20(4^{2x})=20((2^2{)}^{2x})=20(2^{4x})$
   • $2^{x-3}\times 8^{x+2}=2^{x-3}\times (2^3{)}^{x+2}=2^{x-3}\times 2^{3x+6}=2^{(x-3)+(3x+6)}=2^{4x+3}=8\cdot 2^{4x}$

2. Combine terms in numerator.

   $$\text{Numerator}=16\cdot 2^{4x}+20\cdot 2^{4x}=(16+20)\cdot 2^{4x}=36\cdot 2^{4x}$$

3. Simplify the quotient.

   $$\frac{36\cdot 2^{4x}}{8\cdot 2^{4x}}=\frac{36}{8}=\frac{9}{2}$$

Solution

1. Evaluate each exponential term separately.

   • $(64{)}^{-\frac{2}{3}}=(4^3{)}^{-\frac{2}{3}}=4^{3\times (-\frac{2}{3})}=4^{-2}=\frac{1}{4^2}=\frac{1}{16}$
   • $(9{)}^{-\frac{3}{2}}=(3^2{)}^{-\frac{3}{2}}=3^{2\times (-\frac{3}{2})}=3^{-3}=\frac{1}{3^3}=\frac{1}{27}$

2. Perform division.

   $$(64{)}^{-\frac{2}{3}}÷(9{)}^{-\frac{3}{2}}=\frac{1}{16}÷\frac{1}{27}=\frac{1}{16}\times \frac{27}{1}=\frac{27}{16}$$

Solution

1. Express $9$ as $3^2$.

   • $9^{n+1}=(3^2{)}^{n+1}=3^{2(n+1)}=3^{2n+2}$
   • $9^{n-1}=(3^2{)}^{n-1}=3^{2(n-1)}=3^{2n-2}$

2. Combine exponents in numerator and denominator.

   $$\text{Numerator}=3^n\times 3^{2n+2}=3^{n+2n+2}=3^{3n+2}$$ $$\text{Denominator}=3^{n-1}\times 3^{2n-2}=3^{(n-1)+(2n-2)}=3^{3n-3}$$

3. Apply quotient law $\frac{a^m}{a^n}=a^{m-n}$.

   $$\frac{3^{3n+2}}{3^{3n-3}}=3^{(3n+2)-(3n-3)}=3^{3n+2-3n+3}=3^5$$

4. Calculate the final value.

   $$3^5=3\times 3\times 3\times 3\times 3=243$$

Solution

1. Split the exponents in the numerator to isolate $5^n$.

   • $5^{n+3}=5^n\cdot 5^3=125\cdot 5^n$
   • $6\cdot 5^{n+1}=6\cdot (5^n\cdot 5^1)=30\cdot 5^n$

2. Factor out $5^n$ from numerator and denominator.

   $$\text{Numerator}=5^n(125-30)=5^n(95)$$ $$\text{Denominator}=5^n(9-4)=5^n(5)$$

3. Cancel $5^n$ and simplify the fraction.

   $$\frac{5^n\times 95}{5^n\times 5}=\frac{95}{5}=19$$

Solution

1. Calculate $\frac{1}{x}$ by rationalizing.

   Given $x=3+\sqrt{8}$:

   $$\frac{1}{x}=\frac{1}{3+\sqrt{8}}\times \frac{3-\sqrt{8}}{3-\sqrt{8}}=\frac{3-\sqrt{8}}{3^2-(\sqrt{8}{)}^2}=\frac{3-\sqrt{8}}{9-8}=3-\sqrt{8}$$

2. Compute $x+\frac{1}{x}$.

   $$x+\frac{1}{x}=(3+\sqrt{8})+(3-\sqrt{8})=3+\sqrt{8}+3-\sqrt{8}=6$$

Solution

1. Find the value of $\frac{1}{x}$ by rationalizing the denominator.

   Given:

   $$x=3+\sqrt{8}$$

   Taking the reciprocal:

   $$\frac{1}{x}=\frac{1}{3+\sqrt{8}}$$

   Multiplying the numerator and denominator by the conjugate $(3-\sqrt{8})$:

   $$\begin{array}{cc}\frac{1}{x}& =\frac{1}{3+\sqrt{8}}\times \frac{3-\sqrt{8}}{3-\sqrt{8}}\\ & =\frac{3-\sqrt{8}}{(3{)}^2-(\sqrt{8}{)}^2}\\ & =\frac{3-\sqrt{8}}{9-8}\\ & =\frac{3-\sqrt{8}}{1}\\ & =3-\sqrt{8}\end{array}$$

2. Substitute $x$ and $\frac{1}{x}$ into $x-\frac{1}{x}$.

   $$\begin{array}{cc}x-\frac{1}{x}& =(3+\sqrt{8})-(3-\sqrt{8})\\ & =3+\sqrt{8}-3+\sqrt{8}\\ & =(3-3)+(\sqrt{8}+\sqrt{8})\\ & =2\sqrt{8}\end{array}$$

3. Simplify the radical expression.

   Since $\sqrt{8}=\sqrt{4\times 2}=2\sqrt{2}$:

   $$\begin{array}{cc}x-\frac{1}{x}& =2(2\sqrt{2})\\ & =4\sqrt{2}\end{array}$$

Solution

1. Calculate $\frac{1}{x}$ by rationalizing the denominator.

   Given $x=3+\sqrt{8}$:

   $$\frac{1}{x}=\frac{1}{3+\sqrt{8}}\times \frac{3-\sqrt{8}}{3-\sqrt{8}}=\frac{3-\sqrt{8}}{3^2-(\sqrt{8}{)}^2}=\frac{3-\sqrt{8}}{9-8}=3-\sqrt{8}$$

2. Find the sum $x+\frac{1}{x}$.

   $$x+\frac{1}{x}=(3+\sqrt{8})+(3-\sqrt{8})=3+3=6$$

3. Take the square on both sides using $(a+b{)}^2=a^2+2ab+b^2$.

   Squaring both sides of $x+\frac{1}{x}=6$:

   $${(x+\frac{1}{x})}^2=(6{)}^2$$

   Expanding the left hand side:

   $$x^2+2(x)\left(\frac{1}{x}\right)+{\left(\frac{1}{x}\right)}^2=36$$

   Since $x\times \frac{1}{x}=1$:

   $$x^2+2(1)+\frac{1}{x^2}=36$$ $$x^2+2+\frac{1}{x^2}=36$$

4. Subtract $2$ from both sides to isolate $x^2+\frac{1}{x^2}$.

   $$x^2+\frac{1}{x^2}=36-2=34$$

Solution

1. State the algebraic difference of two squares identity.

   Using the algebraic formula:

   $$a^2-b^2=(a+b)(a-b)$$

   Setting $a=x$ and $b=\frac{1}{x}$:

   $$x^2-\frac{1}{x^2}=(x+\frac{1}{x})(x-\frac{1}{x})$$

2. Evaluate the individual factors $(x+\frac{1}{x})$ and $(x-\frac{1}{x})$.

   Given $x=3+\sqrt{8}$ and $\frac{1}{x}=3-\sqrt{8}$:

   $$\begin{array}{cc}x+\frac{1}{x}& =(3+\sqrt{8})+(3-\sqrt{8})=3+3+\sqrt{8}-\sqrt{8}=6\\ x-\frac{1}{x}& =(3+\sqrt{8})-(3-\sqrt{8})=3-3+\sqrt{8}+\sqrt{8}=2\sqrt{8}=4\sqrt{2}\end{array}$$

3. Substitute the values into the difference of squares formula.

   $$\begin{array}{cc}{x}^2-\frac{1}{x^2}& =(x+\frac{1}{x})(x-\frac{1}{x})\\ & =(6)\times (4\sqrt{2})\\ & =24\sqrt{2}\end{array}$$

   (Alternatively: $6\times 2\sqrt{8}=12\sqrt{8}=12(2\sqrt{2})=24\sqrt{2}$.)

Solution

1. Find $x^2+\frac{1}{x^2}$ using the algebraic squaring identity.

   Recall the identity:

   $${(x+\frac{1}{x})}^2=x^2+2(x)\left(\frac{1}{x}\right)+\frac{1}{x^2}=x^2+\frac{1}{x^2}+2$$

   Rearranging to solve for $x^2+\frac{1}{x^2}$:

   $$x^2+\frac{1}{x^2}={(x+\frac{1}{x})}^2-2$$

   Substituting $x+\frac{1}{x}=6$:

   $$\begin{array}{cc}{x}^2+\frac{1}{x^2}& =(6{)}^2-2\\ & =36-2\\ & =34\end{array}$$

2. Apply the squaring formula to $(x^2+\frac{1}{x^2})$ to reach the 4th power.

   Using $(a+b{)}^2=a^2+2ab+b^2$ for $a=x^2$ and $b=\frac{1}{x^2}$:

   $$\begin{array}{cc}{(x^2+\frac{1}{x^2})}^2& =(x^2{)}^2+2(x^2)\left(\frac{1}{x^2}\right)+{\left(\frac{1}{x^2}\right)}^2\\ & =x^4+2+\frac{1}{x^4}\end{array}$$

   Rearranging for $x^4+\frac{1}{x^4}$:

   $$x^4+\frac{1}{x^4}={(x^2+\frac{1}{x^2})}^2-2$$

3. Substitute $x^2+\frac{1}{x^2}=34$ and simplify.

   $$\begin{array}{cc}{x}^4+\frac{1}{x^4}& =(34{)}^2-2\\ & =1156-2\\ & =1154\end{array}$$

Solution

1. Method 1 (Using the algebraic expansion formula).

   Recall the expansion formula:

   $$(a-b{)}^2=a^2-2ab+b^2$$

   For $a=x$ and $b=\frac{1}{x}$:

   $$\begin{array}{cc}{(x-\frac{1}{x})}^2& =x^2-2(x)\left(\frac{1}{x}\right)+\frac{1}{x^2}\\ & =(x^2+\frac{1}{x^2})-2\end{array}$$

   Since $x^2+\frac{1}{x^2}=34$:

   $$\begin{array}{cc}{(x-\frac{1}{x})}^2& =34-2\\ & =32\end{array}$$

2. Method 2 (Direct squaring of the evaluated linear term).

   From $x-\frac{1}{x}=2\sqrt{8}=4\sqrt{2}$:

   $$\begin{array}{cc}{(x-\frac{1}{x})}^2& =(4\sqrt{2}{)}^2\\ & =4^2\times (\sqrt{2}{)}^2\\ & =16\times 2\\ & =32\end{array}$$

   Both algebraic approaches verify that the result is $\mathrm{𝟑𝟐}$.

Solution

1. Rationalize the Left Hand Side (L.H.S).

   Multiply numerator and denominator by the conjugate $4-3\sqrt{2}$:

   $$\text{L.H.S}=\frac{8-3\sqrt{2}}{4+3\sqrt{2}}\times \frac{4-3\sqrt{2}}{4-3\sqrt{2}}=\frac{(8-3\sqrt{2})(4-3\sqrt{2})}{(4+3\sqrt{2})(4-3\sqrt{2})}$$

2. Expand numerator and denominator.

   $$\text{Numerator}=8(4)-8(3\sqrt{2})-3\sqrt{2}(4)+(-3\sqrt{2})(-3\sqrt{2})$$ $$=32-24\sqrt{2}-12\sqrt{2}+9(2)=32-36\sqrt{2}+18=50-36\sqrt{2}$$ $$\text{Denominator}=4^2-(3\sqrt{2}{)}^2=16-9(2)=16-18=-2$$

3. Divide each term by $-2$.

   $$\frac{50-36\sqrt{2}}{-2}=\frac{50}{-2}-\frac{36\sqrt{2}}{-2}=-25+18\sqrt{2}$$

4. Equate corresponding rational and radical parts.

   Given $\text{L.H.S}=p+q\sqrt{2}$:

   $$-25+18\sqrt{2}=p+q\sqrt{2}$$

   Equating rational parts and coefficients of $\sqrt{2}$:

   $$p=-25,q=18$$

Solution

1. Simplify each factor in prime exponential form.

   • $(25{)}^{\frac{3}{2}}=(5^2{)}^{\frac{3}{2}}=5^{2\times \frac{3}{2}}=5^3=125$
   • $(243{)}^{\frac{3}{5}}=(3^5{)}^{\frac{3}{5}}=3^{5\times \frac{3}{5}}=3^3=27$
   • $(16{)}^{\frac{5}{4}}=(2^4{)}^{\frac{5}{4}}=2^{4\times \frac{5}{4}}=2^5=32$
   • $(8{)}^{\frac{4}{3}}=(2^3{)}^{\frac{4}{3}}=2^{3\times \frac{4}{3}}=2^4=16$

2. Substitute simplified values into the expression.

   $$\frac{125\times 27}{32\times 16}$$

3. Multiply out numerator and denominator.

   $$\text{Numerator}=125\times 27=3375$$ $$\text{Denominator}=32\times 16=512$$

4. Final irreducible fraction.

   $$=\frac{3375}{512}$$

Solution

1. Express all terms with base $3$.

   • $\sqrt[3]{(27{)}^{2x}}=\sqrt[3]{(3^3{)}^{2x}}=\sqrt[3]{3^{6x}}=(3^{6x}{)}^{\frac{1}{3}}=3^{2x}$
   • $9^{x+1}=(3^2{)}^{x+1}=3^{2x+2}=3^{2x}\cdot 3^2=9\cdot 3^{2x}$
   • $216(3^{2x-1})=216\left(\frac{3^{2x}}{3^1}\right)=\frac{216}{3}\cdot 3^{2x}=72\cdot 3^{2x}$

2. Combine denominator terms.

   $$\text{Denominator}=9\cdot 3^{2x}+72\cdot 3^{2x}=(9+72)\cdot 3^{2x}=81\cdot 3^{2x}$$

3. Form the quotient and cancel $3^{2x}$.

   $$\frac{54\times 3^{2x}}{81\times 3^{2x}}=\frac{54}{81}$$

4. Reduce fraction by dividing numerator and denominator by $27$.

   $$\frac{54÷27}{81÷27}=\frac{2}{3}$$

Solution

1. Simplify each factor inside the square root.

   • $(216{)}^{\frac{2}{3}}=(6^3{)}^{\frac{2}{3}}=6^{3\times \frac{2}{3}}=6^2=36$
   • $(25{)}^{\frac{1}{2}}=\sqrt{25}=5$
   • $(0.04{)}^{-\frac{3}{2}}={\left(\frac{4}{100}\right)}^{-\frac{3}{2}}={\left(\frac{1}{25}\right)}^{-\frac{3}{2}}=(25{)}^{\frac{3}{2}}=(5^2{)}^{\frac{3}{2}}=5^3=125$

2. Substitute values inside the square root.

   $$\sqrt{\frac{36\times 5}{125}}$$

3. Simplify fraction inside the root.

   $$\frac{36\times 5}{125}=\frac{36}{25}$$

4. Take the square root.

   $$\sqrt{\frac{36}{25}}=\frac{\sqrt{36}}{\sqrt{25}}=\frac{6}{5}$$

Solution

1. Identify the algebraic pattern.

   Recall the algebraic identity:

   $$(x+y)(x^2-xy+y^2)=x^3+y^3$$

2. Set substitutions.

   Let:

   • $x=a^{\frac{1}{3}}⟹x^2={\left(a^{\frac{1}{3}}\right)}^2=a^{\frac{2}{3}}$
   • $y=b^{\frac{2}{3}}⟹y^2={\left(b^{\frac{2}{3}}\right)}^2=b^{\frac{4}{3}}$
   • $xy=a^{\frac{1}{3}}{b}^{\frac{2}{3}}$

3. Apply the identity.

   $$(a^{\frac{1}{3}}+b^{\frac{2}{3}})(a^{\frac{2}{3}}-a^{\frac{1}{3}}{b}^{\frac{2}{3}}+b^{\frac{4}{3}})={\left(a^{\frac{1}{3}}\right)}^3+{\left(b^{\frac{2}{3}}\right)}^3$$

4. Simplify powers.

   $$=a^{3\times \frac{1}{3}}+b^{3\times \frac{2}{3}}=a^1+b^2=a+b^2$$