If a, b, c are real numbers such that a+b+c=0 | vedclass

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Question

(c)

We have, $a+b+c=0$ and $a^2+b^2+c^2=1$

Now $(3 a+5 b-8 c)^2+(-8 a+3 b+5 c)^2 +(5 a-8 b+3 c)^2$

$=9 a^2+25 b^2+64 c^2-48 a c+30 a b -

Vedclass · Accepted Answer

$147$

Vedclass · Answer

(c)

We have, $a+b+c=0$ and $a^2+b^2+c^2=1$

Now $(3 a+5 b-8 c)^2+(-8 a+3 b+5 c)^2 +(5 a-8 b+3 c)^2$

$=9 a^2+25 b^2+64 c^2-48 a c+30 a b -80 b c+64 a^2+9 b^2+25 c^2-80 a c -48 a b+30 b c+25 a^2+64 b^2+9 c^2 +30 a c-8 a b-48 b c$

$=98\left(a^2+b^2+c^2ight)-98(a b+b c+c a)$

$=98\left(a^2+b^2+c^2ight) -98\left(\frac{(a+b+c)^2-\left(a^2+b^2+c^2ight)}{	ext { 2 }}ight)$

$=98(1)-98\left(\frac{0-1}{2}ight)=98+49=147$

If $a, b, c$ are real numbers such that $a+b+c=0$ and $a^2+b^2+c^2=1$, then $(3 a+5 b-8 c)^2+(-8 a+3 b+5 c)^2$ $+(5 a-8 b+3 c)^2$ is equal to

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