Question

If `a+b+c=7`, `a^2+b^2+c^2=35` and `a^3+b^3+c^3=151`, then what is `abc` ?

Answer 1

`(abc)=-15`

Explanation:

`a+b+c=7`
`5+3+(-1)=7`

`a^2+b^2+c^2=35`
`5^2+3^2+(-1)^2`
`=25+9+1`
`=35`

`a^3+b^3+c^3=151`
`5^3+3^3+(-1)^3`
`=125+27-1`
`=151`

`(abc)`
`=5×3×-1`
`=-15`

Answer 2

`abc = -15`

Explanation:

Given:

`{ (a+b+c=7), (a^2+b^2+c^2=35), (a^3+b^3+c^3=151) :}`

Note that:

`(a+b+c)^3`

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

`(a+b+c)(a^2+b^2+c^2)`

`= a^3+b^3+c^3+ab^2+a^2b+bc^2+b^2c+ca^2+c^2a`

So:

`6abc = (a+b+c)^3-3(a+b+c)(a^2+b^2+c^2)+2(a^3+b^3+c^3)`

`color(white)(6abc) = 7^3-3(7)(35)+2(151)`

`color(white)(6abc) = 343-735+302`

`color(white)(6abc) = -90`

Dividing both sides by `6` we find:

`abc = -15`

Answer 3

`-15.`

Explanation:

We will use the following Result :

`a^3+b^3+c^3-3abc`

`=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)...............(star)`

Now, `a+b+c=7 rArr (a+b+c)^2=49.`

`rArr a^2+b^2+c^2+2(ab+bc+ca)=49.`

But, we are given that, `a^2+b^2+c^2=35.`

`:. 35+2(ab+bc+ca)=49, or,`

`ab+bc+ca=(49-35)/2=7...(ast)`

Thus, altogether, we have,

`a^3+b^3+c^3=151, a+b+c=7, a^2+b^2+c^2=35..."(given),"`

`&, by (ast), ab+bc+ca=7.`

Using all these in `(star),` we get,

`151-3abc=(7){35-(7)}=196.`

`:. abc=1/3(151-196)=-45/3=-15,` as deived by,

Respcted George C. Sir.

Enjoy Maths.!