Question

Q : Verify that x^3 + y^3 + z^3 – 3xyz = 1/2 ( x + y + z ) [ ( x – y )^2 + ( y – z)^2 + ( z – x )^2 ]

Class : 9th
Chapter : Polynomials

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Answer 1

$\\ \large \maltese \: \underbrace{ \pmb{ \sf Solution : - }}$

$\\ \quad \qquad \dag \: \underline{ \sf LHS : - }$

$\\ \bullet \quad \sf {x}^{3} + {y}^{3} + {z}^{3} - 3xyz$

$\\ \quad \qquad \dag \: \underline{ \sf RHS : - }$

$\\ \bullet \quad \sf \frac{1}{2} (x + y + z) \bigg[{(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2} \bigg]$

Let's Solve the RHS part by using the following two identities :-

$\\ \leadsto \underline{ \boxed{ \sf{(a - b)}^{2} = {a}^{2} + {b}^{2} - 2ab}} \bigstar$

$\\ \leadsto \underline{ \boxed{ \sf {x}^{3} + {y}^{3} + {z}^{3} = (x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx) }} \bigstar$

Let's get started !!

$\\ \rightarrow \sf \frac{1}{2} (x + y + z) \bigg[ {(x - y)}^{2} + {(y - z)}^{2} + {(z - x)}^{2} \bigg]$

$\\ \rightarrow \sf \frac{1}{2} (x + y + z) \bigg[ ( {x}^{2} + {y}^{2} - 2xy) + ( {y}^{2} + {z}^{2} - 2yz) + ( {z}^{2} + {x}^{2} - 2xz)\bigg]$

$\\ \rightarrow\sf \frac{1}{2} (x + y + z) \bigg[ {x}^{2} + {y}^{2} - 2xy + {y}^{2} + {z}^{2} - 2yz + {z}^{2} + {x}^{2} - 2zx\bigg]$

$\\ \rightarrow \sf \frac{1}{2} (x + y + z) \bigg[ 2 {x}^{2} + 2 {y}^{2} + 2 {z}^{2} - 2xy - 2yz - 2zx \bigg]$

$\\ \rightarrow \sf \frac{1}{2} (x + y + z) \times 2 \bigg [{x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx ) \bigg]$

$\\ \rightarrow \sf (x + y + z) \bigg[ {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx \bigg]$

$\\ \rightarrow \boxed{\sf {x}^{3} + {y}^{3} + {z}^{3} - 3xyz } \bigstar$

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