factorize
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the answer can be further factorised by putting identity of. (x-y)^3
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x³-6x²+12x-8
= (x-2)[x²-4x+4]
= (x-2)[x²-2x-2x+4]
= (x-2)[x(x-2)-2(x-2)]
= (x-2)(x-2)(x-2)
_____________________
onother method
using identity
a³-b³-3ab(a-b) = (a-b)³
x³-6x²+12x-8
= (x)³-(2)³-3(x)(2)[x-2]
= (x-2)³
= (x-2)(x-2)(x-2)
______________________
= (x-2)[x²-4x+4]
= (x-2)[x²-2x-2x+4]
= (x-2)[x(x-2)-2(x-2)]
= (x-2)(x-2)(x-2)
_____________________
onother method
using identity
a³-b³-3ab(a-b) = (a-b)³
x³-6x²+12x-8
= (x)³-(2)³-3(x)(2)[x-2]
= (x-2)³
= (x-2)(x-2)(x-2)
______________________
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