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Find the value of x 3 + y 3 + z 3 – 3xyz if x 2 + y 2 + z 2 = 83 and x + y + z = 15
Dear StudentConsider 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 × 12 x^3+ y^3+ z^3– 3xyz = 180Thanks
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
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