Chapter 12: Q 24. (page 836)

Expand ${(x^{2} - y^{2})}^{6}$ using Binomial Theorem

Short Answer

Expert verified

The value of ${(x^{2} - y^{2})}^{6} = x^{12} - 6 x^{10} y^{2} + 15 x^{8} y^{4} - 20 x^{6} y^{6} + 15 x^{4} y^{8} - 6 x^{2} y^{10} + y^{12}$

Step by step solution

Step 1. Defination

Let $x$ and $a$ be real numbers for any positive integer $n$ , we have ${(x + a)}^{n} = (\begin{matrix} n \\ 0 \end{matrix}) x^{n} + (\begin{matrix} n \\ 1 \end{matrix}) a x^{n - 1} + \dots \dots + (\begin{matrix} n \\ j \end{matrix}) a^{j} x^{n - j} + \dots \dots + (\begin{matrix} n \\ n \end{matrix}) a^{n} = \sum_{j = 0}^{n} (\begin{matrix} n \\ j \end{matrix}) x^{n - j} a^{j}$

Step 2. Explanation

$\Rightarrow {(x^{2} - y^{2})}^{6} = (\begin{matrix} 6 \\ 0 \end{matrix}) {(- y^{2})}^{0} {(x^{2})}^{6} + (\begin{matrix} 6 \\ 1 \end{matrix}) {(- y^{2})}^{1} {(x^{2})}^{5} + (\begin{matrix} 6 \\ 2 \end{matrix}) {(- y^{2})}^{2} {(x^{2})}^{4} + (\begin{matrix} 6 \\ 3 \end{matrix}) {(- y^{2})}^{3} {(x^{2})}^{3} + (\begin{matrix} 6 \\ 4 \end{matrix}) {(- y^{2})}^{4} {(x^{2})}^{2} + (\begin{matrix} 6 \\ 5 \end{matrix}) {(- y^{2})}^{5} {(x^{2})}^{1} + (\begin{matrix} 6 \\ 6 \end{matrix}) {(- y^{2})}^{6} {(x^{2})}^{0}$

$\Rightarrow {(x^{2} - y^{2})}^{6} = x^{12} - 6 x^{10} y^{2} + 15 x^{8} y^{4} - 20 x^{6} y^{6} + 15 x^{4} y^{8} - 6 x^{2} y^{10} + y^{12}$

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Expand ${(x^{2} - y^{2})}^{6}$ using Binomial Theorem

Short Answer

Step by step solution

Step 1. Defination

Step 2. Explanation

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Expand x2-y26using Binomial Theorem

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Step by step solution

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Expand ${(x^{2} - y^{2})}^{6}$ using Binomial Theorem