if a + b + c= 5, ab + bc + ca= 10, Prove that :-a3+ b3 + сз-abc=-252

1.	if a + b + c= 5, ab + bc + ca= 10, Prove that :-a3+ b3 + сз-abc=-252
Answer» 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//

Answer»

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//

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