Question

verify that x³+y³+z3- 3xyz=½ {(x-y)2+(y-z)²+(z-x²)} polynomials ​

Answer 1

✍️Question:-

$\red\bigstar$ Verify, that $\tt x^{3} + y^{3} + z^{3} - 3xyz = \dfrac{1}{2} {(x - y)^{2} + (y - z)^{2} + (z - x)^{2}}$

$\rule{300}{1}$

✍️How to find?

• Using appropriate identity firstly put the values and solve till the identity matches the L.H.S .hence the required answer will come.

✍️Solution:-

☃️Solving R.H.S

$\implies \tt \dfrac{1}{2} (x + y + z) [(x - y)^{2} + (y - z)^{2} + (z - x)^{2}] \\ \tt \Using \: identity (a - b)^{2} = a^{2} + b^{2} - 2ab \\\\ = \tt \dfrac{1}{2} (x + y + z) [(x^{2} + y^{2} + 2xy) + (y^{2} + z^{2} - 2yz) + (z^{2} + x^{2} - 2zx)] \\\\ =\tt \dfrac{1}{2} (x + y + z) [ 2x^{2} + 2y^{2} + 2z^{2} - 2xy - 2yz - 2zx] \\\\ = \tt \dfrac{1}{2} 2[x^{2} + y^{2} + z^{2} - xy -yz -zx] \\\\ =\tt (x + y + z) [x^{2} + y^{2} + z^{2} - xy - yz - zx] \\\\ \tt \We \: know\: x^{3} + y^{3} +z^{3} - 3xyz = (x + y + z) (x^{2} + y^{2} + z^{2} - xy - yz - zx) \\\\ = \tt x^{3} + y^{3} + z^{3} - 3xyz$

= L.H.S.

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Hence, we are completed with our problem.