Ex.2.5 Q14 Polynomials Solution - NCERT Maths Class 9

Go back to  'Ex.2.5'

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,

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