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Identity (x-y)^2

Prove thata2-...

Question

Prove that

(a²-b²)³+(b²-c²)³ +(c²-a²)³ divided by

(a-b)³ +(b-c)³+(c-a)^3.equals to. (a+b)(b+c)(c+a)

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Solution

---> if A+ B+ C = 0 ; A³ + B³ + C³ = 3ABC
Using above formula
( a² - b² ) + ( b² - c² ) + ( c² - a² ) = 0
therefore
( a² - b² )³ + ( b² - c² )³ + ( c² - a² )³ = 3( a² - b² )( b² - c² )( c² - a² )
( a - b ) + ( b - c ) + ( c - a ) = 0
( a - b )³ + ( b - c )³ + ( c - a )³ = 3( a - b )( b - c )( c - a )
[ ( a² - b² )³ + ( b² - c² )³ + ( c² - a² )³ divided by ( a - b )³ + ( b - c )³ + ( c - a )³
=3( a² - b² )( b² - c² )( c² - a² ) divided by 3( a - b )( b - c )( c - a )
= 3(a-b)(a+b)(b-c)(b+c) (c-a)(c+a) divided by 2(a-b)(b-c)(c-a)
= (a+b)(b+c)(c+a)

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Similar questions

\frac{(a^{2} - b^{2})^{3} + (b^{2} - c^{2})^{3} + (c^{2} - a^{2})^{3}}{(a - b)^{3} + (b - c)^{3} + (c - a)^{3}}

\frac{(a^{2} - b^{2})^{3} + (b^{2} - c^{2})^{3} + (c^{2} - a^{2})^{3}}{(a - b)^{3} + (b - c)^{3} + (c - a)^{3}}

\frac{(a^{2} - b^{2})^{3} + (b^{2} - c^{2})^{3} + (c^{2} - a^{2})^{3}}{(a - b)^{3} + (b - c)^{3} + (c - a)^{3}} =

(a) 3(a + b) ( b+ c) (c + a)

(b) 3(a − b) (b − c) (c − a)

(c) (a − b) (b − c) (c − a)

(d) none of these

Q. The value of

3 [\frac{(a^{2} - b^{2})^{3} + (b^{2} - c^{2})^{3} + (c^{2} - 1^{2})^{3}}{(a - b)^{3} + (b - c)^{3} + (c - a)^{3}}]

Q. (a − b)³ + (b − c)³ + (c − a)³ =

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

(b) (a − b) (b − c) (c − a)

(c) 3(a − b) ( b− c) (c − a)

(d) none of these

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