Factorize 1 + 64a3 using suitable identity:
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Given:
= > 1 + 64a^3
= > 1^3 + (4a)^3
We know that a^3 + b^3 = (a + b)(a^2 - ab + b^2)
= > (4a + 1)((1)^2 - 1 * 4a + (4a)^2)
= > (4a + 1)(1 - 4a + 16a^2)
= > (4a + 1)(16a^2 - 4a + 1).
Hope this helps!
= > 1 + 64a^3
= > 1^3 + (4a)^3
We know that a^3 + b^3 = (a + b)(a^2 - ab + b^2)
= > (4a + 1)((1)^2 - 1 * 4a + (4a)^2)
= > (4a + 1)(1 - 4a + 16a^2)
= > (4a + 1)(16a^2 - 4a + 1).
Hope this helps!
siddhartharao77:
:-)
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