Question

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

Answer 1

Answer:

Step-by-step explanation:

we know that

a^3+b^3+c^3 - 3abc= (a+b+c)(a^2+b^2+c^2- 2ab-2bc-2ca)

as, a+b+c= 5 and ab+bc+ca= 10

and a^2+b^2+c^2= (a+b+c)^2 - 2(ab+bc+ca)

so rhs= 5[(5)^2 - 2(10)- 2(ab+bc+ca)]

=5[ 25-20-20]= -75.

u check the question nicely as the answer should be -75

Answer 2

Step-by-step explanation:

a+ b+ c = 5

ab + bc + ca = 10

(a+ b + c )^2 = a^2 + b ^2 + c^2 + 2 ( ab + bc + ca )

(5)^2 - 2 × 10 = a ^2 + b^2 + c ^2

a ^2 + b^2 + c ^2 = 5

Now ,

a^3 + b ^3 + c^3 - 3abc = (a+ b+ c)( a^2 + b ^2 + c^2 - ab - bc - ca )

(5) (5-10) = 5×-5

= -25

Hence a3+ b3 + c3-3abc =-25 proved