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  • Expand the following : (i)(3a-2b)^{3} (ii)\left ( \frac{1}{x}+\frac{y}{3} \right )^{3} (iii)\left ( 4-\frac{1}{3x} \right )^{3}

Expand the following :

(i)(3a-2b)^{3}

(ii)\left ( \frac{1}{x}+\frac{y}{3} \right )^{3}

(iii)\left ( 4-\frac{1}{3x} \right )^{3}

 

Answers (1)

(i)27a^{3}-8b^{2}-54a^{2}b+36ab^{2}

Solution: Given (3a-2b)^{3}

We know that

(a-b)^{3}=a^{3}-b^{3}-3a^{2}b+3ab^{2}

So we get:

(3a-2b)^{3}=(3a)^{3}-(2b)^{3}-3(3a)^{2}(2b)+3(3a)(2b)^{2}

(3a-2b)^{3}=27a^{3}-8b^{2}-54a^{2}b+36ab^{2}

Hence the expanded form is 27a^{3}-8b^{2}-54a^{2}b+36ab^{2}

 

(ii)\frac{1}{x^{3}}+\frac{y^{3}}{27}+\frac{y}{x^{2}}+\frac{y^{2}}{3x}

Solution: Given \left ( \frac{1}{x}+\frac{y}{3} \right )^{3}

We know that

(a+b)^{3}=a^{3}+b^{3}+3a^{2}b+3ab^{2}

So we get:

\left ( \frac{1}{x}+\frac{y}{3} \right )^{3}=\left (\frac{1}{x} \right )^{3}+\left (\frac{y}{3} \right )^{3}+3 \left (\frac{1}{x} \right )^{2}\left (\frac{y}{3} \right )+3\left ( \frac{1}{x} \right )\left (\frac{y}{3} \right )^{2}

=\frac{1}{x^{3}}+\frac{y^{3}}{27}+\frac{3y}{3x^{2}}+\frac{3y^{2}}{9x}

=\frac{1}{x^{3}}+\frac{y^{3}}{27}+\frac{y}{x^{2}}+\frac{y^{2}}{3x}

Hence the expanded form is =\frac{1}{x^{3}}+\frac{y^{3}}{27}+\frac{y}{x^{2}}+\frac{y^{2}}{3x}

 

(iii)64-\frac{1}{27x^{3}}-\frac{16}{x}+\frac{4}{3x^{2}}

Solution: Given \left ( 4-\frac{1}{3x} \right )^{3}

We know that

(a-b)^{3}=a^{3}-b^{3}-3a^{2}b+3ab^{2}

So we get:

\left ( 4-\frac{1}{3x} \right )^{3}=(4)^{3}-\left ( \frac{1}{3x} \right )^{3}-3(4)^{2}\left ( \frac{1}{3x} \right )+3(4)\left ( \frac{1}{3x} \right )^{2}
=64-\frac{1}{27x^{3}}-\frac{48}{3x}+\frac{12}{9x^{2}}

=64-\frac{1}{27x^{3}}-\frac{16}{x}+\frac{4}{3x^{2}}

Hence the expanded form is  64-\frac{1}{27x^{3}}-\frac{16}{x}+\frac{4}{3x^{2}}.

Posted by

infoexpert24

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