Question

Find the value of x3 + y3 + z3 – 3xyz if x2 + y2 + z2 = 83 and x + y + z = 15​

Answer 1

Answer:

(a+b+c) = 15

squaring on both side

(a+b+c)² = (15)²

a²+b²+c²+2ab+2bc+2ac = 225

83 + 2(ab+bc+ac) = 225

2(ab+bc+ac) = 225 - 83

2(ab+bc+ac) = 142

ab+bc+ac = 71 ----------- 1

now ,

a³+b³+c³ = 3abc

(a+b+c) (a²+b²+ç²-ab-bc-ac)

15(83- (ab+bc+ac)

15(83 - 71)

15 × 12

180

Answer 2

Answer:

Consider the equation x + y + z = 15

From algebraic identities, we know that (a + b + c)2 = a2 + b2 + c2 + 2(ab + bc + ca)

So,

(x + y + z)2 = x2 + y2 + z2 + 2(xy + yz + xz)

From the question, x2 + y2 + z2 = 83 and x + y + z = 15

So,

152 = 83 + 2(xy + yz + xz)

=> 225 – 83 = 2(xy + yz + xz)

Or, xy + yz + xz = 142/2 = 71

Using algebraic identity a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca),

x3 + y3 + z3 – 3xyz = (x + y + z)(x² + y² + z² – (xy + yz + xz))

Now,

x + y + z = 15, x² + y² + z² = 83 and xy + yz + xz = 71

So, x3 + y3 + z3 – 3xyz = 15(83 – 71)

x3 + y3 + z3 – 3xyz = 15 × 12

Or, x3 + y3 + z3 – 3xyz = 180