Dear Student,(a2-b2)3+(b2-c2)3+(c2-a2)3(a-b)3+(b-c)3+(c-a)3If A + B +C =0 , then A3+ B3+C3 =3ABCWe can apply this formula in both numerator and denominator =3(a2-b2)(b2-c2)(c2-a2)3(a-b)(b-c)(c-a)=(a2-b2)(b2-c2)(c2-a2)(a-b)(b-c)(c-a) [ A2-B2=(A-B)(A+B) ]=(a -b )(a+b)(b-c )(b+c)(c-a)(c+a)(a-b)(b-c)(c-a)=(a -b )(a+b)(b-c )(b+c)(c-a)(c+a)(a-b)(b-c)(c-a)=(a+b)(b+c)(c+a) As option B and D are not visible, please verify the correct option with above result Hope this information will clear your doubts about topic Regards

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Solve 1 question Column equal to NSE S\KVPY/NISE foundation Simplify (a2 —b2)J +(b2 —c2)3 • (ci — ali (a-b)3 • (b - c)) •(c —a) (C) 3abC If p -2— a, value of a' + 6ap• p: —Bis —e Of 303 • 203 - 503 + 90000

D e a r S t u d e n t, \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}} I f A + B + C = 0, t h e n A^{3} + B^{3} + C^{3} = 3 A B C W e c a n a p p l y t h i s f o r m u l a i n b o t h n u m e r a t o r a n d d e n o m i n a t o r . = \frac{3 (a^{2} - b^{2}) (b^{2} - c^{2}) (c^{2} - a^{2})}{3 (a - b) (b - c) (c - a)} = \frac{(a^{2} - b^{2}) (b^{2} - c^{2}) (c^{2} - a^{2})}{(a - b) (b - c) (c - a)} [A^{2} - B^{2} = (A - B) (A + B)] = \frac{(a - b^{}) (a + b) (b - c) (b + c) (c - a) (c + a)}{(a - b) (b - c) (c - a)} = \frac{(a - b^{}) (a + b) (b - c) (b + c) (c - a) (c + a)}{(a - b) (b - c) (c - a)} = (a + b) (b + c) (c + a)

As option B and D are not visible, please verify the correct option with above result.

Hope this information will clear your doubts about topic.

Regards

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Solve 1 question Solve 1 question Column equal to NSE S\KVPY/NISE foundation Simplify (a2 —b2)J +(b2 —c2)3 • (ci — ali (a-b)3 • (b - c)) •(c —a) (C) 3abC If p -2— a, value of a' + 6ap• p: —Bis —e Of 303 • 203 - 503 + 90000

Solve 1 question

Solve 1 question Column equal to NSE S\KVPY/NISE foundation Simplify (a2 —b2)J +(b2 —c2)3 • (ci — ali (a-b)3 • (b - c)) •(c —a) (C) 3abC If p -2— a, value of a' + 6ap• p: —Bis —e Of 303 • 203 - 503 + 90000