Question

if a^6+b^6-c^6+3a^2b^2c^2=0,then prove that a^2+b^2=c^2​

Answer 1

Answer:

$\displaystyle \sf \longrightarrow (a^2+b^2)=(c^2) \ [ \ Proved \ ]$

Step-by-step explanation:

Given :

$\displaystyle \sf a^6+b^6-c^6+3 \ a^2.b^2.c^2=0$

We have to prove :

$\displaystyle \sf a^2+b^2=c^2$

Rewrite the given as :

$\displaystyle \sf a^6+b^6+3 \ a^2.b^2.c^2=c^6$

On solving it further :

$\displaystyle \sf \longrightarrow (a^2)^3+(b^2)^3+3 \ (a^2.b^2).c^2=(c^2)^3$

Putting :

$\displaystyle \sf \longrightarrow c^2=a^2+b^2$

$\displaystyle \sf \longrightarrow (a^2)^3+(b^2)^3+3 \ (a^2.b^2).(a^2+b^2)=(c^2)^3$

Using identity :

$\displaystyle \sf \longrightarrow x^3+y^3+3 \ xy.(x+y)= (x+y)^3$

$\displaystyle \sf \longrightarrow (a^2+b^2)^3=(c^2)^3$

Taking cube root both side we get :

$\displaystyle \sf \longrightarrow \left[(a^2+b^2)^3\right]^{1/3}=\left[(c^2)^3\right]^{1/3}$

$\displaystyle \sf \longrightarrow (a^2+b^2)=(c^2)$

Hence proved.

Answer 2

Answer:

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