Question

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

Answer 1

ANSWER:

• Value of a³+b³+c³-3abc = 108

GIVEN:

• a+b+c = 9
• a²+b²+c² = 35

TO FIND:

• a³+b³+c³-3abc

SOLUTION:

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

Answer 2

ANSWER:

Value of a³+b³+c³-3abc = 108

GIVEN:

a+b+c = 9

a²+b²+c² = 35

TO FIND:

a³+b³+c³-3abc

SOLUTION:

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.