Question

a) Verify the following:
(a? - b2) (a? + b2) + (62 – c2) (b2 + c2) + (c2 – a2) (c2 + a2) = 0​

Answer 1

✈️ Question :-

Verify the following -

$\sf (a^2-b^2)(a^2+b^2)+(b^2-c^2) \\ \sf (b^2+c^2)+(c^2-a^2)(c^2+a^2)=0$

✈️ Answer :-

$\sf (a^2-b^2)(a^2+b^2) \: is \: the \: form \: of: \\ (x^2-y^2)(x^2+y^2)=(x^2-y^2)$

$\sf So, (a^2-b^2)=(a^2-b^2)(a^2+b^2)$

$\sf Similarly, (b^2-c^2)=(b^2-c^2)(b^2+c^2)$

$\sf (c^2-a^2)=(c^2-a^2)(c^2+a^2)$

$\sf (a^2-b^2) + (b^2-c^2) + (c^2-a^2)=RHS$

$\sf 0=LHS$

$\mapsto \sf (a^2-b^2) + (b^2-c^2) + (c^2-a^2)=0$

$\mapsto \sf a^2-b^2+b^2-c^2+c^2-a^2=0$

$\mapsto \sf \cancel{a^2}\cancel{-a^2}\cancel{+b^2}\cancel{-b^2}\cancel{+c^2}\cancel{-c^2}=0$

$\mapsto \sf 0=0$

Hence, proved !

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Hope it helps you :)