Question

If x+y+z=0. Than prove that x³y³z³=3cuz.

Answer 1

Answer :-

→ There is a mistake in the questions it should be

• If x + y + z = 0 , Then prove that x³ + y³ + z³ = 3xyz

◾ Provided that

▪️x + y + z = 0

◾ Asked to prove that

▪️x³ + y³ + z³ = 3xyz

⏩ By using the identity that

→ $\tiny{a^3 + b^3 + c^3 - 3abc = (a + b + c)( a^2 + b^2 + c^2 - ab - ac - bc)}$

◾We can say that

$\implies \tiny{x^3 + y^3 + z^3 - 3xyz = (x + y + z)( x^2 + y^2 + z^2 - xy - xz - yz)}$

▪️ Now by further solving

$\implies \tiny{x^3 + y^3 + z^3 = (x + y + z)( x^2 + y^2 + z^2 - xy - xz - yz) + 3xyz}$

▪️ Now substituting ( x + y + z) = 0

$\implies \tiny{x^3 + y^3 + z^3 = (0)( x^2 + y^2 + z^2 - xy - xz - yz) + 3xyz}$

$\implies \sf{x^3 + y^3 + z^3 = 0 + 3xyz }$

♦ Hence proved that x³ + y³ + z³ = 3xzy