Question

If w is complex cube of unity show that (3+3w+5w^2)^6-(2+6w+2w^2)^3=0

Answer 1

Explanation:

if w is a complex cube of unity, hence it satisfies the equation

${w}^{2} + w + 1 = 0$

${(3 + 3w + 5{w}^{2})}^{6} - {(2 + 6w + 2 {w}^{2})}^{3}$

$= > {(3 + 3w + 3 {w}^{2} + 2 {w}^{2}) }^{6} -$

$(2 + 2w + 2 {w}^{2} + 4w)^{3} = >$

${(2 {w}^{2})}^{6} - {(4w)}^{3} = > 64 {w}^{6} - 64 {w}^{3} = >$

0. Hence proved. Hope it helps you.

Answer 2

Answer:

It is proved that \frac{a + bw + cw^{2} }{aw + bw^{2} + c} = w^{2}aw+bw2+ca+bw+cw2=w2

Step-by-step explanation:

From the property of the imaginary cube root of unity, w

We have, w³ = 1 and w² + w + 1 = 0

Now, we have to prove that  \frac{a + bw + cw^{2} }{aw + bw^{2} + c} = w^{2}aw+bw2+ca+bw+cw2=w2

Now, \frac{a + bw + cw^{2} }{aw + bw^{2} + c}aw+bw2+ca+bw+cw2

= \frac{w^{3} (a + bw + cw^{2}) }{aw + bw^{2} + c}aw+bw2+cw3(a+bw+cw2) {Since, w³ = 1 }

= \frac{(aw^{3} + bw^{4} + cw^{5}) }{(aw + bw^{2} + c)}(aw+bw2+c)(aw3+bw4+cw5) {Multiplying w³ to the numerator}

= w^{2} \frac{(aw^{1} + bw^{2} + cw^{3}) }{(aw + bw^{2} + c)}w2(aw+bw2+c)(aw1+bw2+cw3) {Taking w² as common in the numerator}

= w^{2} \frac{(aw + bw^{2} + c) }{(aw + bw^{2} + c)}w2(aw+bw2+c)(aw+bw2+c) {Since, w³ = 1}

= w² (Proved)