Question

if x=a+b, y=aw² +b w and z =aw+bw² show that x³+y³+z³ where w,w²are cube root unity​

Answer 1

Answer:

Step-by-step explanation:

We have,

$\tt{x=a+b\,\,\,\,,\,\,\,\,y=a\,{\omega}^{2}+b\,\omega\,\,\,\,,\,\,\,\,z=a\,\omega+b\,{\omega}^{2}}$

Now,

$\tt{x+y+z=a+b+a\,{\omega}^{2}+b\,\omega+a\,\omega+b\,{\omega}^{2}}$

$\tt{\implies\,x+y+z=a\left(1+\omega+{\omega}^{2}\right)+b\left(1+\omega+{\omega}^{2}\right)}$

Since ω, ω² are cube roots of unity, so,

1+ω+ω²=0           ...(1)

$\tt{\implies\,x+y+z=0}$

So,

$\tt{\implies\,{x}^{3}+{y}^{3}+{z}^{3}=3\,x\,y\,z}$

$\tt{\implies\,{x}^{3}+{y}^{3}+{z}^{3}=3\left(a+b\right)\left({a}^{2}\,{\omega}^{2}+b\,\omega\right)\left(a\,\omega+b\,{\omega}^{2}\right)}$

$\tt{\implies\,{x}^{3}+{y}^{3}+{z}^{3}=3\left(a+b\right)\left({a}^{2}\,{\omega}^{3}+ab\,{\omega}^{2}+ab\,{\omega}^{4}+{b}^{2}\,{\omega}^{3}\right)}$

we know, ω³=1

$\tt{\implies\,{x}^{3}+{y}^{3}+{z}^{3}=3\left(a+b\right)\left({a}^{2}(1)+ab\,{\omega}^{2}+ab\,{\omega}^{3}\cdot\omega+{b}^{2}(1)\right)}$

$\tt{\implies\,{x}^{3}+{y}^{3}+{z}^{3}=3\left(a+b\right)\left({a}^{2}(1)+ab\,{\omega}^{2}+ab\,(1)\cdot\omega+{b}^{2}(1)\right)}$

$\tt{\implies\,{x}^{3}+{y}^{3}+{z}^{3}=3\left(a+b\right)\left({a}^{2}+ab\,{\omega}^{2}+ab\,\omega+{b}^{2}\right)}$

$\tt{\implies\,{x}^{3}+{y}^{3}+{z}^{3}=3\left(a+b\right)\left\{{a}^{2}+ab\left({\omega}^{2}+\omega\right)+{b}^{2}\right\}}$

From (1), we get,

$\tt{\implies\,{x}^{3}+{y}^{3}+{z}^{3}=3\left(a+b\right)\left\{{a}^{2}+ab\left(-1\right)+{b}^{2}\right\}}$

$\tt{\implies\,{x}^{3}+{y}^{3}+{z}^{3}=3\left(a+b\right)\left({a}^{2}-ab+{b}^{2}\right)}$

$\tt{\implies\,{x}^{3}+{y}^{3}+{z}^{3}=3\left({a}^{3}+{b}^{3}\right)}$