Question

is 1/a²- 1/b² divisible by (a-b)???​

Answer 1

$\frac{\Big( \frac{1}{a^{2} }- \frac{1}{b^{2}}\Big) }{ (a-b) }$

$= \frac{ \frac{(b^{2} - a^{2})}{a^{2} b^{2} }}{ (a-b) }$

$=\frac{(b+a)(b-a)}{a^{2}b^{2} (a-b)}$

$= \frac{-(b+a)\cancel {(a-b)}}{a^{2}b^{2} \cancel {(a-b)}}$

$= \frac{-(a+b)}{a^{2}b^{2}}$

Therefore .,

$\red{ \Big( \frac{1}{a^{2}} - \frac{1}{b^{2}}\Big) } \: is \green { \: divisible \: by \: (a-b ) }$

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Answer 2

${\tt{\purple{\underline{\underline{\huge{Answer:}}}}}}$

$\large{\dfrac{\Big( \dfrac{1}{a^{2} }- \dfrac{1}{b^{2}}\Big) }{ (a-b) }}$

$\sf\implies \dfrac{ \dfrac{(b^{2} - a^{2})}{a^{2} b^{2} }}{ (a-b) }$

$\sf\implies \dfrac{(b+a)(b-a)}{a^{2}b^{2} (a-b)}$

$\sf\implies \dfrac{-(b+a)\cancel {(a-b)}}{a^{2}b^{2} \cancel {(a-b)}}$

$\sf\implies \dfrac{-(a+b)}{a^{2}b^{2}}$

$\large{\boxed{\bf{ \Big( \frac{1}{a^{2}} - \frac{1}{b^{2}}\Big) } \: is \: divisible \: by \: (a-b )}}$

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