factorise x cube + 1 by x cube minus 2
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Answered by
66
Formula:-
a^3+b^3+c^3 -3.a.b.c=(a+b+c)(a^2+b^2+c^2-a.b-bc-ca)………….(1)
x^3 + 1/x^3 - 2
=x^3+1/x^3+1 -3
=(x)^3+(1/x)^3 +(1)^3–3.x.1/x.1
=(x+1/x+1)(x^2+1/x^2+1^2-x.1/x-1/x.1–1.x)
=(x+1/x+1)(x^2+1/x^2+1–1–1/x-x)
=(x+1/x+1)(x^2+1/x^2–1/x-x) , Answer
a^3+b^3+c^3 -3.a.b.c=(a+b+c)(a^2+b^2+c^2-a.b-bc-ca)………….(1)
x^3 + 1/x^3 - 2
=x^3+1/x^3+1 -3
=(x)^3+(1/x)^3 +(1)^3–3.x.1/x.1
=(x+1/x+1)(x^2+1/x^2+1^2-x.1/x-1/x.1–1.x)
=(x+1/x+1)(x^2+1/x^2+1–1–1/x-x)
=(x+1/x+1)(x^2+1/x^2–1/x-x) , Answer
Answered by
26
here is your answer.
Hope you are able to understand it.
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