Question

Prove that:- a³+b³+c³-3abc=(a+b+c) (a²+b²+c²-ab-bc-ca). Please tell how we get this (a+b+c) (a²+b²+c²-ab-bc-ca) from a³+b³+c³-3abc??

Answer 1

To prove :

• a³ + b³ + c³ - 3abc = (a + b + c) (a² + b² + c² - ab - bc - ca)

Proof :

Taking RHS :

→ a² + b² + c² - ab - bc - ca

→ a (a² + b² + c² - ab - bc - ca) + b (a² + b² + c² - ab - bc - ca + c (a² + b² + c² - ab - bc - ca)

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

→ a³ + b³ + c³ + ab² - ab² + ac² - ac² + a²b - a²b + a²c - a²c + bc² - bc² + b²c - b²c - abc - abc - abc

→ a³ + b³ + c³ - 3abc

Which is equal to the LHS

$\sf \: Hence, \: Proved!$

We can find out by this method too :

Taking LHS :

• a³ + b³ + c³ - 3abc

→ ( a³ + b³ ) + c³ - 3abc

→ ( a + b)³ - 3ab( a + b) +c³ - 3abc

→ ( a + b)³ - 3a²b - 3ab² + c³ - 3abc

→ {(a + b)³ + c³ } -3a²b - 3ab² - 3abc

→ ( a + b + c)³ - 3c(a + b) ( a + b + c ) -3ab(a + b+ c)

→ ( a + b + c){ ( a + b + c)² -3c(a +b) - 3ab }

→ ( a + b + c){ a² + b² + c² +2ab +2bc+ 2ca -3ca - 3bc - 3ab }

→ ( a + b + c)( a² + b² + c² - ab - bc -ca)

Which is equal to the RHS

$\sf \: Hence, \: Proved!$

Answer 2

Given :-

✭ a³+b³+c³-3abc = (a+b+c)(a²+b²+c²-ab-bc-ca)

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To Prove :-

◈ The Above statement is true

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Proof :-

Before Solving we should know the formula,

a³ + b³ + c³ - 3abc = (a + b + c)(a² + b² + c² - ab - bc - ca)

Solving LHS

➝ a³ + b³ + c³ - 3abc

➝ (a³ + b³ ) + c³ - 3abc

➝ (a + b)³ - 3ab(a + b) +c³ - 3abc

➝ (a + b)³ - 3a²b - 3ab² + c³ - 3abc

➝ [(a + b)³ + c³] -3a²b - 3ab² - 3abc

➝ (a + b + c)³ - 3c(a + b)(a + b + c ) -3ab(a + b+ c)

➝ (a + b + c)[(a + b + c)² -3c(a +b) - 3ab]

➝ (a + b + c)[a² + b² + c² +2ab +2bc+ 2ca -3ca - 3bc - 3ab]

➝ (a + b + c)(a² + b² + c² - ab - bc -ca)

LHS = RHS

$\sf Hence \ Proved !!$

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