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Choose one of the factors of 1331x3 – 8y3 4 11x – 2y 121x2 + 44xy + 4y2 121x2 – 22xy + 4y2
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Since x^3 is a cube, and 8y^3 is the cube of 2y, this looks like it can be the difference of two cubes. Is 1331 a perfect cube? If it is, then do you know how to factor the difference of two cubes?
Use the identity\[x^3-y^3=(x-y)(x^2+xy+y^2)\]
Here, \(x^3=1331x^3\) & \(y^3=8y^3\). u can now easily find out the factors.
can u ?
11x – 2y
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\[1331 x^3-8 y^3=(11 x-2 y) \left(121 x^2+22 x y+4 y^2\right) \]
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