If a+b+c=5 and ab+bc+ca=10 then prove that a^2+b^2+c^2-3abc=-25
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Answered by
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
0
There is mistake in your question there will be cube in stead of square...
Hope this will help you
Hope this will help you
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