Factorise x³+1/x³-2.
Answers
Answered by
0
x
3
+
x
3
1
−2
x
3
+
x
3
1
+1−3
∵x
3
+y
3
+z
3
−3xyz=(x+y+z)(x
2
+y
2
+z
2
−xy−yz−zx)
⇒x
3
+
x
3
1
+1
3
−3×x×
x
1
×1
=(x+
x
1
+1)(x
2
+
x
2
1
+1−1−
x
1
−x)
=(x+
x
1
+1)(x
2
+
x
2
1
−
x
1
−x)
Answered by
0
x^3+1/x^3-2
=(x^3-1)+(1/x^3-1)
=(x-1)(x^2+x+1)+(1/x-1)(1/x^2+1/x+1) (a^3-b^3)=(a-b)(a^2+ab+b^2)
=(x-1)(x^2+x+1)+(1-x)(1+x+x^2)/x^3
=(x-1)(x^2+x+1)-(x-1)(x^2+x+1)/x^3
=(x-1)(x^2+x+1)(1-1/x^3)
=(x-1)(x^2+x+1)(x^3-1)/x^3
=[{(x-1)(x^2+x+1)}^2]/x^3
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=(x^3-1)+(1/x^3-1)
=(x-1)(x^2+x+1)+(1/x-1)(1/x^2+1/x+1) (a^3-b^3)=(a-b)(a^2+ab+b^2)
=(x-1)(x^2+x+1)+(1-x)(1+x+x^2)/x^3
=(x-1)(x^2+x+1)-(x-1)(x^2+x+1)/x^3
=(x-1)(x^2+x+1)(1-1/x^3)
=(x-1)(x^2+x+1)(x^3-1)/x^3
=[{(x-1)(x^2+x+1)}^2]/x^3
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