Question

If a + b + c = 15 and a 2 + b 2 + c 2 = 83, find the value of a 3 + b 3 + c 3 – 3abc.

Answer 1

Answer:

a³+b³+c³-3abc = 180

Step-by-step explanation:

Given,

a+b+c = 15 ----(1)

a²+b²+c² = 83 ---(2)

We know the algebraic identity:

a²+b²+c²+2(ab+bc+ca) = (a+b+c)²

=> 83 + 2(ab+bc+ca) = 15²

/* from (1) & (2) */

=> 2(ab+bc+ca) = 225 - 83

=> 2(ab+bc+ca) = 142

=> ab+bc+ca = 142/2

=> ab+bc+ca = 71 ----(3)

Now ,

By the algebraic identity :

a³+b³+c³-3abc

= (a+b+c)[a²+b²+c²-(ab+bc+ca)]

= 15(83-71)/*From (1),(2)&(3) */

= 15 × 12

= 180

Therefore,

a³+b³+c³-3abc = 180

Answer 2

Step-by-step explanation:

Given: a + b + c = 15, a² + b² + c² = 83

∴ (a + b + c)² = a² + b² + c² + 2(ab + bc + ca)

⇒ 15² = 83 + 2(ab + bc + ca)

⇒ 225 - 83 = 2(ab + bc + ca)

⇒ 142 = 2(ab + bc + ca)

⇒ ab + bc + ca = 71

Now,

a³ + b³ + c³ - 3abc:

= (a + b + c)(a² + b² + c² - ab - bc - ca)

= (a + b + c)(a² + b² + c² - (ab + bc + ca))

= (15)(83 - 71)

= 180.

Hence, the value of a³+b³+c³-3abc is 180

Hope it helps!