Question

If a+b+c=6;asquare+b square+csquare=14and a cube + bcube + c cube=36then prove that abc=6

Answer 1

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

6^2= 14+2(ab+bc+ac)

36 -14= 2(ab+bc+ac)

22 /2= (ab+bc+ac)

ab+bc+ac=11

now,

a^3+b^3+c^3–3abc=(a+b+c)(a^2+b^2+c^2-(ab+bc+ac))

36–3abc=6(14–11)

36–3abc=18

36–18 =3abc

3abc=18

abc= 6. PROVED

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