Question

solve your questions​

Answer 1

Answer:

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

Step-by-step explanation:

To find--->

-----------

(b²-c²)³+(c²-a²)³+(a²-b²)³

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

(b-c)³+(c-a)³+(a-b)³

Solution--->

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first we find value of numerator

for this let

b²-c²=p , c²-a²=q ,a²-b²=r

now

p+q+r=b²-c² + c²-a²+a²-b²

all terms cancel each other so

p + q + r =0

so

p³ + q³ + r³ = 3 p q r

( b²-c²)³+(c²-a²)³+(a²-b²)³=

3(b²-c²)(c²-a²)(a²-b²)

now cocentrating on denominator

(b-c)³+ (c-a)³+ (a-b)³

let

(b-c)=x ,(c-a)=y,(a-b)=z

x+y+z=b-c+c-a+a-b

terms cancel each other in RHS

x+y+z=0

so

x³+y³+z³=3xyz

(b-c)³+(c-a)³+(a-b)³=3(b-c)(c-a)(a-b)

now returning to original problem

(b²-c²)³+(c²-a²)³+(a²-b²)³

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(b-c)³+(c-a)³+(a-b)³

3(b²-c²)(c²-a²)(a²-b²)

=-----------------------------------

3(b-c)(c-a)(a-b)

now we have an identity u²-v²=(u+v)(u-v)

applying it in numerator

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

=-----------------------------------------------------------

3 (b-c)(c-a)(a-b)

3(b+c)(c+a)(a+b) cancel out from numerator and denominator

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

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

Hope it helps u