# Ex.2.5 Q14 Polynomials Solution - NCERT Maths Class 9

## Question

Without actually calculating the cubes, find the value of each of the following:

(i) \begin{align}(-12)^{3}+(7)^{3}+(5)^{3}\end{align}

(ii) \begin{align}(28)^{3}+(-15)^{3}+(-13)^{3}\end{align}

Video Solution
Polynomials
Ex 2.5 | Question 14

## Text Solution

Reasoning:

If \begin{align}x+y+z=0\end{align} then

\begin{align}x^{3}+y^{3}+z^{3}=3 x y z\end{align}

Steps:

(i) Let \begin{align}x&=-12, y=7, z=5 \end{align}

Then\begin{align}x+y+z=-12+7+5=0\end{align}

So by using the identity,

\begin{align}(-12)^{3}+(7)^{3}+(5)^{3} &=3(-12)(7)(5) \\ &=-1260 \end{align}

(ii)  Let $$x$$ \begin{align}&=28, y=-15, z=-13 \end{align}

Then \begin{align}x+y+z=28-15-13=0 \end{align}

So by using the identity,

\begin{align}&(28)^{3}+(-15)^{3}+(-13)^{3} \\&=3(28)(-15)(-13) \\ &=16380 \end{align}

Video Solution
Polynomials
Ex 2.5 | Question 14
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