Let ω≠$$1$$, be a cube root of unity, and a,b∈R.$$Statement-1$$: $$a3+b3=(a+b)a$$ω$$+b$$ω$$2a$$ω$$2+b$$ω$$Statement-2$$: $$x3$$−$$1=(x$$−$$1)x$$ω$$2$$−ωxω−ω$$2$$ for each x∈R.

Question

Let ω≠$$1$$, be a cube root of unity, and a,b∈R.$$Statement-1$$: $$a3+b3=(a+b)a$$ω$$+b$$ω$$2a$$ω$$2+b$$ω$$Statement-2$$: $$x3$$−$$1=(x$$−$$1)x$$ω$$2$$−ωxω−ω$$2$$ for each x∈R.

Infinity Learn · Accepted Answer

We have –xxω$$2$$−ωxω−ω$$2$$    $$=x2$$ω$$3$$−xω$$2$$−xω$$4+$$ω$$3$$    $$=x2$$−ω$$2+$$ω$$x+1$$     $$=x2+x+1$$ ∵ω$$2+$$ω$$=$$−$$1$$∴ (x$$−$$1)x$$ω$$2$$−ωxω−ω$$2$$                         $$=(x$$−$$1)x2+x+1=x3$$−$$1Thus$$, $$Statement-2$$ is true. Replacing x by –x, we $$get($$−$$x)3$$−$$1=($$−x−$$1)−xω$$2$$−ω−xω−ω$$2$$⇒      $$x3+1=(x+1)x$$ω$$2+$$ωxω$$+$$ω$$2Taking$$ conjugate of both the sides, we get        $$x3+1=(x+1)x$$ω$$+$$ω$$2x$$ω$$2+$$ω                                                           [∴  ω$$=$$ω$$2$$]If $$b=0$$, , $$statement-1$$ is clearly true. Suppose b≠$$0$$.Replacing x by $$a/b$$ we get      $$ab3+1=ab+1abw+w2abw2+w$$⇒ $$a3+b3=(a+b)aw+bw2aw2+bwThus$$, $$statement-1$$ is also true and $$statement-2$$ is a correct explanation for it.

Let $ω \neq 1,$ be a cube root of unity, and $a, b \in R$ .
Statement $- 1$ : $a^{3} + b^{3} = (a + b) (a ω + b ω^{2}) (a ω^{2} + b ω)$
Statement $- 2$ : $x^{3} - 1 = (x - 1) (x ω^{2} - ω) (x ω - ω^{2})$ for each $x \in R .$

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Let ω≠1, be a cube root of unity, and a,b∈R.Statement-1: a3+b3=(a+b)aω+bω2aω2+bωStatement-2: x3−1=(x−1)xω2−ωxω−ω2 for each x∈R.

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Let $ω \neq 1,$ be a cube root of unity, and $a, b \in R$ .
Statement $- 1$ : $a^{3} + b^{3} = (a + b) (a ω + b ω^{2}) (a ω^{2} + b ω)$
Statement $- 2$ : $x^{3} - 1 = (x - 1) (x ω^{2} - ω) (x ω - ω^{2})$ for each $x \in R .$