Question

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

ANSWER WITH WORKING THX

Answer 1

Start with

`(a+b+c)^3=a^3+b^3+c^3+3(a^2b+a^2c+ab^2+ac^2+b^2c+bc^2)+6abc`

Given that
`a+b+c=5`, we know
`(a+b+c)^3=125`.

Also,
`ab+bc+ca=10`, so

`a^2b+a^2c+ab^2+ac^2+b^2c+bc^2=a(ab+ac)+b(ab+bc)+c(ac+bc)`

`=a(10-bc)+b(10-ca)+c(10-ab)`

`=10(a+b+c)-3abc`

`=50-3abc`

So we have

`125=a^3+b^3+c^3+3(50-3abc)+6abc`

`125=a^3+b^3+c^3+150-9abc+6abc`

`\implies -25=a^3+b^3+c^3-3abc`

QED