how to factor x^4-y^4?
and also factor 16a^4 - 81b^4 which is related
I know difference of two cubes is
x^3+y^3 = (x+y)(x^2-xy+y^2)
x^3-y^3 = (x-y)(x^2+xy+y^2)
and difference of two squares is
x^2-y^2 = (x-y)(x+y)
6 Answers
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We treat it just like the difference of two squares, since (x^2)^2 = x^4.
x^4 - y^4 = (x^2 + y^2)(x^2 - y^2)
But, you'll notice the final term is ALSO a difference of squares, so we repeat:
=(x^2+y^2)(x+y)(x-y)
Hope that helps!
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For X4 processors, there are not much differences... they all have four CPU's, they only differ I guess by speed. But by experience, I'm using Intel Quad Core Q6600 2.4GHz with a 4GB RAM and it worked great! with a Windows Vista Home Basic.
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(X^4 - Y^4) = (X^2 - Y^2)(X^2 + Y^2)
(16a^4 - 81b^4) = (4a^2 - 9b^2)(4a^2 + 9b^2)
Hope that helps.
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Same way that you do squares.
(x^2+y^2)(x^2-y^2)=x^4-y^4
(4a^2+9b^2)(4a^2-9b^2)=16a^4-81b^4
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and the sum of two squares is:
x^2 + y^2 = (x - yi)(x + yi)
(x^4 - y^4)
(x^2 - y^2) (x^2 + y^2)
(x - y)(x + y)(x - yi)(x + yi)
16a^4 - 81b^4
(4a^2 - 9b^2) (4a^2 + 9b^2)
(2a - 3b) (2a + 3b) (2a - 3bi) (2a + 3bi)
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(x²-y²)(x²+y²) =
(x+y)(x-y)(x²+y²)
16a^4 - 81b^4 =
(4a²+9b²)(4a²-9b²)=
(4a²+9b²)(2a+3b)(2a-3b).
.....