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प्रश्न
Factorise the following, using the identity a2 – 2ab + b2 = (a – b)2.
a2y3 – 2aby2 + b2y
योग
उत्तर
We have,
a2y3 – 2aby2 + b2y = y(a2y2 – 2aby + b2)
= y[(ay)2 – 2 · ay · b + b2]
= y(ay – b)2
= y(ay – b)(ay – b)
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अध्याय 7: Algebraic Expression, Identities and Factorisation - Exercise [पृष्ठ २३५]
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