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  • Without actually calculating the cubes, find the value of (i) \left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{3} \right )^{3}-\left ( \frac{5}{6} \right )^{3} (ii) (0.2)^{3}-(0.3)^{3}+(0.1)^{3}

Without actually calculating the cubes, find the value of

(i) \left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{3} \right )^{3}-\left ( \frac{5}{6} \right )^{3}

(ii) (0.2)^{3}-(0.3)^{3}+(0.1)^{3}

Answers (1)

(i) \frac{-5}{12}

Given \left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{3} \right )^{3}-\left ( \frac{5}{6} \right )^{3}

We know that

a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

When a + b + c = 0, we get a^{3}+b^{3}+c^{3}=3abc

Here,

\frac{1}{2} +\frac{1}{3}- \frac{5}{6} =\frac{3+2-5}{6}=\frac{0}{6}=0

So using the above identity, we get:

\therefore \left ( \frac{1}{2} \right )^{3}+\left ( \frac{1}{3} \right )^{3}+\left ( \frac{-5}{6} \right )^{3}=3 \times \frac{1}{2} \times \frac{1}{3}\times \frac{-5}{6}=\frac{-5}{12}

Hence the answer is \frac{-5}{12}

(ii)-0.018

Solution

Given: (0.2)^{3}-(0.3)^{3}+(0.1)^{3}

We know that

a^{3}+b^{3}+c^{3}-3abc=(a+b+c)(a^{2}+b^{2}+c^{2}-ab-bc-ca)

When a + b + c = 0, we get a^{3}+b^{3}+c^{3}=3abc

Here,

0.2+(-0.3)+0.1=0

So,                              
(0.2)^{3}-(0.3)^{3}+(0.1)^{3}=3(0.2)(-0.3)(0.1)=-0.018

Hence the answer is -0.018.

Posted by

infoexpert24

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