Question

if a+ b+ c = 10 and ab + bc + ca = 20 the value of a3 + b3 + c3 - 3abc​

Answer 1

Step-by-step explanation:

Given,

a+b+c=10 and ab+bc+ca=20

So, to find the value of a3+b3+c3-3abc is-

By using the identity,

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

Now we need to find the value of a2+b2+c2

We also know,

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

Putting the value of a+b+c=10 and ab+bc+ca=20

$(10)^{2} = a^{2} +b^{2} + c^{2} + 2(20)$

=> $100=a^{2} + b^{2} +c^{2} + 40$

=> $a^{2} + b^{2} + c^{2} = 100-40$

=>$a^{2} +b^{2} +c^{2} = 60$

Now,

Substituting the value in the formula for a³ + b³ + c³- 3abc

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

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

a³ + b³ + c³- 3abc = ( 10 ) ( 60 - ( 20 ) )

a³ + b³ + c³- 3abc =  ( 10 ) ( 40 )

a³ + b³ + c³- 3abc = 400

Hence,

a³ + b³ + c³- 3abc = 400 (Ans)