Question

a³+b³+c³-3abc = a+b+c/2 ((a-b)²(b-c)²(c-a)²)​

Answer 1

Question: Prove that a³ + b³ + c³ - 3abc = $\frac{1}{2}$(a+b+c) [(a-b)² + (b-c)² +(c-a)²]

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Answer

a³ + b³ + c³ - 3abc =  (a+b+c) [(a-b)² + (b-c)² +(c-a)²]

We first write down the LHS, And expand it

$\frac{1}{2}$ (a+b+c) [(a-b)² + (b-c)² +(c-a)²]

$\frac{1}{2}$ (a+b+c) [a² + b² - 2ab + b² + c² - 2bc + c² + a² - 2ca]

Clubbing common terms and seperating them

$\frac{1}{2}$ (a+b+c) [2a² + 2b² + 2c²] [-2ab - 2bc - 2ca]

Since 2 is common on both brackets, we take it out.

$\frac{a + b + c}{2}$ 2 {a² + b² + c²] [-ab - bc - ca]  

(2 and 2 get cancelled)

{a + b + c} {a² + b² + c²] [-ab - bc - ca]

Now we multiply a, b, and c with the Highlighted terms.  

a {a² + b² + c²] [-ab - bc - ca] + b {a² + b² + c²] [-ab - bc - ca] + c {a² + b² + c²] [-ab - bc - ca]

After multiplying this is the result. [All the bolded terms get cancelled]

a³ + ab² + ac² - a²b - abc - ca² + ba² + b³ + bc² - ab² - b²c - abc + ca² + cb² + c³ - cab - bc² - c²a

Now we write down the left out terms.

a³ - abc + b³ - bac + c³ - cab

a³ + b³ + c³ - abc - bac - cab

[Abc is repeated 3 times}

a³ + b³ + c³ - 3abc

LHS = RHS

Hence Proved :D

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