Question

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


(Solution is *108*. To be the BRAINLIEST provide right and detailed steps)

Answer 1

Given that:

• a + b + c = 9

• a²+b²+c²= 35

To find:

• The value of a³+b³ +c³-3abc.

Solution:

• Using 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

Therefore, the value of a³+b³ +c³-3abc is 108.