62x^2+7n-3 factorise
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(4m-7m)[16m²+28mn+49n²]
We know the algebraic identity:
a³-b³ = (a-b)(a²+ab+b²)
Given
64m³ - 343n³
= (4m)³ - (7n)³
= (4m-7m)[(4m)²+(4m)(7n)+(7n)²]
= (4m-7m)[16m²+28mn+49n²]
Therefore,
64m³-343n³
= (4m-7m)[16m²+28mn+49n²]
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