Question

Verify
that a3 + b3+c3-3abc= 1/2(a+b+c)[(a-b)²+(b-c) whole square+(c-a) whole square​

Answer 1

Answer:

By taking RHS and then equate it with LHS

Answer 2

$\mathbb{ELLO \ THERE!}$

------------------

Question: Prove that a³ + b³ + c³ - 3abc$= \frac{1}{2}$

(a+b+c) [(a-b)² + (b-c)² +(c-a)²]

------------------

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

--------------------------------

$\mathbb{THANKS!}$