If a+b+c=0 and ab+bc+ca=40, find a^2+b^2+c^2
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2
Answer:
-80
Step-by-step explanation:
We know that the tes given can be compared to the identity which says that
-(a+b+c) ^2=a^2+b^2+c^2+2(ab+bc+ca)
Since, a+b+c=0 and
ab+bc+can=40
Therefore : (0)^2=a^2+b^2+c^2+2(40)
{I have inserted 0 and 40 bcoz it is mentioned that a+b+c=0 and ab + bc+ ca=40}
=> 0 =a^2+b^2+c^2+80
=> 0-80=a^2+b^2+c^2
=> -80=a^2+b^2+c^2
Answer is -80
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