Question

if x+y+z=15 x²+y²+z²=33 find the value of x^3+y^3+z^3-3xyz

Answer 1

Answer:

x³ + y³ + z³ - 3xyz = -945

Step-by-step explanation:

We know,

x + y + z = 15 _______(1)

x² + y² + z² = 33 _______(2)

Cube both sides in equation (1):

= (x + y + z)³ = 15³

= (x + y + z)(x² + y² + z² + 2xy + 2xz + 2yz) = 3375 _______(3)

Using equation (1) and (2) in equation (3):

= 15(33 + 2(xy + xz + yz)) = 3375

= 495 + 30(xy + xz + yz) = 3375

Dividing whole equation by 15:

= 33 + 2(xy + xz + yz) = 225

= 2(xy + xz + yz) = 225 - 33

= xy + xz + yz = 192 ÷ 2

= xy + xz + yz = 96 ________(4)

Also from equation (1):-

x + y = 15 - z _______(5)

y + z = 15 - x _______(6)

x + z = 15 - y _______(7)

Now, using equation (4) in equation (3):

= x³ + y³ + z³ + xy² + xz² + x²y + yz² + x²z + zy² + 192(x + y + z) = 3375

Using equation (1):

= x³ + y³ + z³ + xy² + xz² + + x²y + yz² + x²z + zy² + 192 × 15 = 3375

= x³ + y³ + z³ + xy² + xz² + + x²y + yz² + x²z + zy² = 3375 - 2880

= x³ + y³ + z³ + xy² + x²z + xz² + x²y + yz² + zy² = 495

= x³ + y³ + z³ + xy² + x²y + x²z + xz² + yz² + y²z = 495

= x³ + y³ + z³ + xy(y + x) + xz(x + z) + yz(z + y) = 495

Using equations (5), (6) and (7):

= x³ + y³ + z³ + xy(15 - z) + xz(15 - y) + yz(15 - x) = 495

= x³ + y³ + z³ + 15xy - xyz + 15xz - xyz + 15yz - xyz = 495

= x³ + y³ + z³ + 15(xy + xz + yz) - 3xyz = 495

= x³ + y³ + z³ - 3xyz + 1440 = 495

= x³ + y³ + z³ - 3xyz = 495 - 1440

= x³ + y³ + z³ - 3xyz = -945