If | a |=a and | b |=b, prove that ( a a^2 - b b^2 )^2= ( a - b ab )^2.

( a a^2 - b b^2 )^2 = | a a^2 |^2+ | b b^2 |^2= 2 a . b a^2b^2 = | a |^2 a^4 + | b |^2 b^4 - 2 a . b a^2b^2 = a^2 a^4 + b^2 b^4 - 2 a . b a^2b^2 (From the given information) =1 a^2 +1 b^2 - 2 a . b a^2b^2 = b^2+a^2-2 a . b a^2b^2 = a^2+b^2-2 a . b a^2b^2 = | a |^2+| a |^2-2 a . b a^2b^2 (From the given information) = ( a - b )^2 a^2b^2 = ( a - b ab )^2

If | a |=a and | b |=b, prove that ( a a^2 - b b^2 )^2= ( a - b a…

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Question

If $|\vec{\text{a}}|=\text{a}$ and $\big|\vec{\text{b}}\big|=\text{b},$ prove that $\Big(\frac{\vec{\text{a}}}{\text{a}^2}-\frac{\vec{\text{b}}}{\text{b}^2}\Big)^2=\Big(\frac{\vec{\text{a}}-\vec{\text{b}}}{\text{ab}}\Big)^2.$

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Answer

$\Big(\frac{\vec{\text{a}}}{\text{a}^2}-\frac{\vec{\text{b}}}{\text{b}^2}\Big)^2$
$=\Big|\frac{\vec{\text{a}}}{\text{a}^2}\Big|^2+\Big|\frac{\vec{\text{b}}}{\text{b}^2}\Big|^2=\frac{2\vec{\text{a}}.\vec{\text{b}}}{\text{a}^2\text{b}^2}$
$=\frac{|\vec{\text{}a}|^2}{\text{a}^4}+\frac{\big|\vec{\text{b}}\big|^2}{\text{b}^4}-\frac{2\vec{\text{a}}.\vec{\text{b}}}{\text{a}^2\text{b}^2}$
$=\frac{\text{a}^2}{\text{a}^4}+\frac{\text{b}^2}{\text{b}^4}-\frac{2\vec{\text{a}}.\vec{\text{b}}}{\text{a}^2\text{b}^2}$ (From the given information)
$=\frac{1}{\text{a}^2}+\frac{1}{\text{b}^2}-\frac{2\vec{\text{a}}.\vec{\text{b}}}{\text{a}^2\text{b}^2}$
$=\frac{\text{b}^2+\text{a}^2-2\vec{\text{a}}.\vec{\text{b}}}{\text{a}^2\text{b}^2}$
$=\frac{\text{a}^2+\text{b}^2-2\vec{\text{a}}.\vec{\text{b}}}{\text{a}^2\text{b}^2}$
$=\frac{|\vec{\text{a}}|^2+|\vec{\text{a}}|^2-2\vec{\text{a}}.\vec{\text{b}}}{\text{a}^2\text{b}^2}$ (From the given information)
$=\frac{\big(\vec{\text{a}}-\vec{\text{b}}\big)^2}{\text{a}^2\text{b}^2}$
$=\Big(\frac{\vec{\text{a}}-\vec{\text{b}}}{\text{ab}}\Big)^2$