Math, asked by SohanSr8939, 1 year ago

If a+b+c=5 and ab+bc+ca=10 then prove that a^2+b^2+c^2-3abc=-25

Answers

Answered by sonabrainly
1

we know ,


a³ + b³ + c³ -3abc = (a + b + c )(a² + b² + c² -ab -bc-ca)


now ,


a + b + c = 5

ab + bc + ca = 10


(a + b + c)² = a² + b² + c² +2(ab + bc+ca)

(5)² -2×10 = a² + b² + c²

a² + b² + c² =5


hence ,


a³ + b³ +c³ -3abc = ( a + b + c )(a² + b² + c² -ab- bc-ca)


=( 5)( 5 - 10) = 5 × (-5) = -25


hence proved//




Answered by Tamash
0
There is mistake in your question there will be cube in stead of square...

Hope this will help you
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