$({a-b)}^{2} {+—=a}^{2} {-b}^{2}$

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Solution

Explanation:
We have to find the value of the blank . Let $— = x$
It means we find the value of $x$ .
We know that
$({a-b)}^{2} {=(a-b)(a-b)=a}^{2} {-ab-ba+b}^{2} (aftermultiplyingthebrackets) = a^{2} {-ab-ab+b}^{2} (ab=ba) = a^{2} {-2ab+b}^{2}$
So $({a-b)}^{2} {+x=a}^{2} {-b}^{2} {x=a}^{2} {-b}^{2} {-(a-b)}^{2} {x=a}^{2} {-b}^{2} {-(a}^{2} {-2ab+b}^{2}) {x=a}^{2} {-b}^{2} {-a}^{2} {+2ab-b}^{2} (duetominussignthesignofthetermswithinthebracketchanges) {x=2ab-2b}^{2} {(herea}^{2} iscancelledby)$
Hence, ${2ab-b}^{2}$ is the required answer.

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\frac{a}{b}

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Which of the following is correct?
a) $(a - b)^{2} = a^{2} + 2 a b - b^{2}$
b) $(a - b)^{2} = a^{2} - 2 a b + b^{2}$
c) $(a - b)^{2} = a^{2} - b^{2}$
d) $(a + b)^{2} = a^{2} + 2 a b - b^{2}$

If A = [\begin{matrix} a & b b & a \end{matrix}] and A^{2} = [\begin{matrix} α & β β & α \end{matrix}], then

\frac{\sqrt{a^{2} - b^{2}} + a}{\sqrt{a^{2} + b^{2}} + b} \div \frac{\sqrt{a^{2} + b^{2}} - b}{a - \sqrt{a^{2} - b^{2}}}

Q. Simplify:

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(a-b)2+—=a2-b2

$({a-b)}^{2} {+—=a}^{2} {-b}^{2}$