Question

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Solve this with well explanations . ✍️✍️

$\large\bold\red{Q.}$ If $\bold{x = a + b, y =a \omega + b { \omega}^{2} }$ and $\bold{z = a { \omega }^{2} + b \omega}$, then which one of the Following is true :-

$(A) {x}^{3} + {y}^{3} + {z}^{3} = 3( {a}^{3} + {b}^{3} ) \\ \\ (B)xyz = 2( {a}^{3} + {b}^{3} ) \\ \\ (C)x + y + z \ne 0 \\ \\ (D) {x}^{2} + {y}^{2} + {z}^{2} = {a}^{2} + {b}^{2}$​

Answer 1

Hey there!

Given,

x = a + b

y = aω + bω²

z = aω² + bω

Then,

x + y + z = a + b + aω + bω² + aω² + bω

              = a + aω + aω² + b + bω + bω²

             = (a+b)(1 +  ω + ω²)

             = (a + b) * 0

x + y + z = 0    (i.e Option (C) is wrong)

Let's check option (A);

Since x + y + z = 0

Then, x³ + y³ + z³ = 3xyz .......(i)

Now Finding value of xyz:

xyz

= (a+b) (aω + bω²) ( aω² +bω)

= (a+b) (a²ω³ + abω² + abω⁴ + b²ω³)

= (a+b) ω³(a² + abω² +abω + b²)

= (a+b){a²+ab(ω² + ω) + b²}     {∵ω³ = 1}

= (a+b)(a² +ab(-1) + b²)           {∵ ω² + ω = -1}

= (a+b) (a² - ab + b²)

= a³ + b³

Putting value of xyz in equation (i);

x³ + y³ + z³ = 3xyz

x³ + y³ + z³ = 3(a³ + b³)

Hence, option(A) is correct.

As you've mentioned that one option is correct, so we don't have the need to check other options.

Answer 2

Answer:

Step-by-step explanation: