Ex.2.5 Q7 Polynomials Solution - NCERT Maths Class 9

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Question

Evaluate the following using suitable identities:

(i) \(\begin{align}(99)^{3}\end{align}\)

(ii) \(\begin{align}(102)^{3} \end{align}\)

(iii) \(\begin{align}(998)^{3} \end{align}\)

 Video Solution
Polynomials
Ex 2.5 | Question 7

Text Solution

  

Reasoning:

\(\begin{align}{\bf{Identities:}} (x+y)^{3}=x^{3}+y^{3}+3 x y(x+y)\\{(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y)}\end{align}\)

Steps:

(i)\(\begin{align}\;\;(99)^{3}=(100-1)^{3}\end{align}\)

Identity: \(\begin{align}(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y)\end{align}\)

Take \(\begin{align}x=100, y=1\end{align}\)

\[\begin{align}(99)^{3} &=(100)^{3}-(1)^{3}-3(100)(1)(100-1) \\ &=1000000-1-300 \times 99 \\ &=999999-29700 \\ &=9,70,299 \end{align}\]

(ii)\(\begin{align}\;\;(102)^{3}=(100+2)^{3}\end{align}\)

Identity:  \(\begin{align}(x+y)^{3}=x^{3}+y^{3}+3 x y(x+y)\end{align}\)

Take \(\begin{align}x=100, y=2\end{align}\)

\[\begin{align}(102)^{3} &=(100)^{3}+(2)^{3}+3(100)(2)(100+2) \\ &=1000000+8+600 \times 102 \\ &=1000008+61200 \\ &=10,61,208 \end{align}\]

(iii)\(\begin{align}\;\;(998)^{3}=(1000-2)^{3}\end{align}\)

Identity: \(\begin{align}(x-y)^{3}=x^{3}-y^{3}-3 x y(x-y)\end{align}\)

Take \(\begin{align}x=1000, y=-2\end{align}\)

\[\begin{align}(998)^{3} &=(1000)^{3}-(2)^{3}-3(1000)(2)(1000-2) \\ &=1000000000-8-6000\,\times\,998 \\ &=999999992+5988000 \\ &=99,40,11,992 \end{align}\]