# 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:

Identities:

\begin{align} (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}

Video Solution
Polynomials
Ex 2.5 | Question 7

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