If `a+b+c=6` and `a b+b c+c a=11` , find the value of `a^3+b^3+c^3-3a b cdot`

Explanation

we know that
` a^(3) + b^(3) + c^(3) - 3abc = (a+b+c) (a^(2)+b^(2) +c^(2) - ab - bc -ca )`
`= a^(3) + b^(3) + c^(3) - 3 abc = ( a+ b+ c) { (a^(2) + b^(2) + c^(2) ) - (ab+ bc + ca)}" " ` …(i)
Clearly: To find the value of ` a^(3) + b^(3) + c^(3) - 3abc ` `a+bc,a^(2)+b^(2)+c^(2)` and `ab+bc+ca`, `a+b+c` , we know the values of `ab+bc+ca` , Therefore, we will find value of `a^(2) + b^(2)+c^(2)`
We know that
`(a+b+c)^(2) = a^(2) + b^(2) + c^(2)+ 2ab + 2bc + 2ca `
` rArr ( a+ b+ c)^(2) = (a^(2) + b^(2) + c^(2) ) + 2 (ab+bc+ ca)`
` rArr 6^(2) = a^(2) + b^(2) + c^(2) + 2 xx 11`'' " [ Putting values of `a+ b+ c`" and " `ab + bc + ca" "]`
` rArr 36 = a^(2) + b^(2) + c^(2) + 22 `
` rArr a^(2) + b^(2) + c^(2) = 36 - 22 `
` rArr a^(2) + b^(2) + c^(2) = 14 `
In equation ( i) `a+b+c = 6 , ab+ bc + ca = 1 ` and ` a^(2) + b^(2) + c^(2) `
` a^(3) + b^(3) + c^(3) - 3abc = 6 xx (14-11) = 6 xx 3 = 18 `