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Factor binomial: 64x^3-1. How do i do this with a x^3 I dont understand :(
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\[64x^{3}-1\]
Hi starshine7! Here we can use the formula, \[a ^{3}-b ^{3}=(a-b)(a ^{2}+ab+b ^{2})\] hence, \[(8x)^{3}-(1)^{3}\] Now can you do it by yourself?
It is \[ (4x)^3 - (1)^3 \]
\[(4x-1)(\ (4x)^2+(4x)(1)+(1)^2) \] \[(4x-1)(16x^2+4x+1)\]
oh yes 4 not 8. thanx diyadiya
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np :)
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