Question

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

Answer 1

SOLUTION :

Consider the equation x + y + z = 15

From algebraic identities, we know that (a + b + c)³ = 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 .

Answer 2

Hola mate

Here is your answer -

SOLUTION :

Consider the equation x + y + z = 15

From algebraic identities, we know that (a + b + c)³ = 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 .