Question

FACTORIZE: The equation given above in the picture. Please solve it quickly.

Answer 1

Solution :

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We know that ,

If x + y + z = 0 then

x³ + y³ + z³ = 3xyz

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i ) Let x = a²-b² , y = b²-c² , z = c² - a²

x + y + z

= a² - b² + b² - c² + c² - a²

= 0

Therefore ,

(a²-b²)³+(b²-c²)³+(c²-a²)³

= 3(a²-b²)(b²-c²)(c²-a²) ----( 1 )

ii ) Similarly ,

(a-b)³+(b-c)³+(c-a)³ = 3(a-b)(b-c)(c-a)---(2)

Now ,

[(a²-b²)³+(b²-c²)³+(c²-a²)³]/[(a-b)³+(b-c)³+(c-a)³]

= [3(a²-b²)(b²-c²)(c²-a²)]/[3(a-b)(b-c)(c-a)]

= (a+b)(b+c)(c-a)

[ Since , x² - y² = ( x + y )( x - y ) ]

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