Question

if x=1,y= - 2 and z=3, find the value of x³+ y³+z³- 3xyz​

Answer 1

Consider the equation x + y + z = 15From algebraic identities, we know that (a + b + c)^2= a^2+ b^2+ c^2+ 2(ab + bc + ca)So,(x + y + z)^2= x^2+ y^2+ z^2+ 2(xy + yz + xz)From the question, x^2+ y^2+ z^2= 83 and x + y + z = 15So,152= 83 + 2(xy + yz + xz)=> 225 – 83 = 2(xy + yz + xz)Or, xy + yz + xz = 142/2 = 71Using algebraic identity a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca),x^3+ y^3+ z^3– 3xyz = (x + y + z)(x² + y² + z² – (xy + yz + xz))Now,x + y + z = 15, x² + y² + z² = 83 and xy + yz + xz = 71So, x^3+ y^3+ z^3– 3xyz = 15(83 – 71)x^3+ y^3+ z^3– 3xyz = 15 × 12x^3+ y^3+ z^3– 3xyz = 180

Answer 2

Answer:

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