Question

If a+b+c=9,a²+b²+c²=35 find the value of a³+b³+c³-3abc

Answer 1

(a+b+c)²=9²=81
a²+b²+c²+2ab+2bc+2ca=81
a²+b²+c²+2ab+2bc+2ca-(a²+b²+c²)=81-35
2ab+2bc+2ca=46
2(ab+bc+ca)=46
(ab+bc+ca)=23
a³+b³+c³-3abc=(a+b+c)(a²+b²+c²-ab-bc-ca)
=9(35-(ab+bc+ca))
=9(35-23)
=9(12)=108

Answer 2

Formula:

=> a³+b³+c³-3abc = (a+b+c)[a²+b²+c²-(ab+bc+ca)]

Here:

a+b+c = 9

a²+b²+c² = 35

=> a+b+c=9

Squaring both sides we get:

=> (a+b+c)² = (9)²

=> a²+b²+c²+2(ab+bc+ca) = 81

Putting a²+b²+c² = 35

=> 35 +2(ab+bc+ca) = 81

=> 2(ab+bc+ca) = 81-35

=> ab+bc+ca = 46/2

=> ab+bc+ca = 23

=> a³+b³+c³-3abc = (a+b+c)[a²+b²+c²-(ab+bc+ca)]

Putting values in R.H.S we get;

= 9(35-23)

= 9(12)

= 108

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