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

3 Answers

(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

Apr 17, 2017

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

Apr 17, 2017

-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.!