If `a+b+c=9` and `a^2+b^2+c^2=35 ,` find the value of `a^3+b^3+c^3-3a b c`

Explanation

Given:
`a+b+c=9` and `a^2+b^2+c^2=35`
Now, `a+b+c=9`
squaring both the sides,
`=>``(a+b+c)^2=9^2`
`=>``a^2+b^2+c^2+2(ab+bc+ac)=81`
`=>``2(ab+bc+ca)=81−35 [a^2+b^2+c^2=35]`
`=>ab+bc+ca=46/2=23`
`x^3+y^3+z^3−3xyz=(x+y+z)[x^2+y^2+z^2−(xy+yz+xz)]`
`a^3+b^3+c^3−3abc=(a+b+c)[a^2+b^2+c^2−(ab+bc+ca)]`
`a^3+b^3+c^3−3abc=(9)[35−(23)]`
`=>``a^3+b^3+c^3−3abc=9xx12`
`=>``a^3+b^3+c^3−3abc=108`
Hence, `a^3+b^3+c^3−3abc=108`