Question

If a+b+c=5 and ab+bc+ca=10 then prove that a3+b3+c3-3abc= -25 .

Answer 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//

Answer 2

a+b+c=5
ab+bc+ca=10

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

Finding a²+b²+c²,
(a + b + c)² = a² + b² + c² +2(ab+bc+ca)
(5)² -2×10 = a² + b² + c²
a²+b²+c² =5

a³+b³+c³-3abc=(5)(5-(ab+bc+ca)
a³+b³+c³-3abc=5(5-10)
=5(-5)
=-25

Hence proved