how do I factor out 64-8x^3?
回答 (6)
2008-11-23 6:57 am
✔ 最佳答案
a³-b³ = (a-b)(a²+ab+b²)64-8x³
= 8(8-x³)
= 8(2³-x³)
= 8(2-x)(2²+2x+x²)
= 8(2-x)(4+2x+x²)
2008-11-23 4:26 pm
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
64 - 8x^3
= 8(8 - x^3)
= 8(2^3 - x^3)
= 8(2 - x)(2^2 + 2x + x^2)
= 8(2 - x)(4 + 2x + x^2)
64 - 8x^3
= 8(8 - x^3)
= 8(2^3 - x^3)
= 8(2 - x)(2^2 + 2x + x^2)
= 8(2 - x)(4 + 2x + x^2)
2008-11-23 3:04 pm
both terms are divisible by "8", so you can pull it out of both. that would leave you with
8(8 - x^3)
you can factor further because of the difference of two cubes, leaving the final answer of
8(2 - x)(x^2 + 2x + 4)
8(8 - x^3)
you can factor further because of the difference of two cubes, leaving the final answer of
8(2 - x)(x^2 + 2x + 4)
2008-11-23 3:02 pm
(4-2x)(16+8x+4x^2)=
=64+32x+16x^2 -32x-16x^2-8x^3=
=64-8x^3
=2(2-x)(4)(4+2x+x^2)=
=8(2-x)(x^2+2x+4)
=64+32x+16x^2 -32x-16x^2-8x^3=
=64-8x^3
=2(2-x)(4)(4+2x+x^2)=
=8(2-x)(x^2+2x+4)
2008-11-23 2:58 pm
rewrite it as 4^3 - (2x)^3 = (4-2x)(16 + 8x + 4x^2)
note: a^3 - b^3 = (a-b)(a^2 + ab + b^2)
note: a^3 - b^3 = (a-b)(a^2 + ab + b^2)
2008-11-23 2:57 pm
64 = 4^3
8 = 2^3
64 - 8x^3 = 4^3 - 2^3 x^3 = 4^3 - (2x)^3
Now, use the difference of cubes identity:
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
a=4, b=2x
64 - 8x^3 = (4 - 2x)(16 + 8x + 4x^2)
8 = 2^3
64 - 8x^3 = 4^3 - 2^3 x^3 = 4^3 - (2x)^3
Now, use the difference of cubes identity:
a^3 - b^3 = (a - b)(a^2 + ab + b^2)
a=4, b=2x
64 - 8x^3 = (4 - 2x)(16 + 8x + 4x^2)
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