Question

Can you please solve this question?Step by step explanation required.If you give the correct answer,I will mark you as the brainliest
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Answer 1

Answer:

1)

a³ + b³ + c³ - 3abc = (a + b + c )(a² + b² + c² -ab - ac -bc)

Now it is given that : a³ + b³ + c³ = 3abc

So,

=> a³ + b³ + c³ - 3abc = 0

=> (a + b + c )(a² + b² + c² -ab - ac -bc) = 0

that is =>>>

(a² + b² + c² -ab - ac -bc) = 0 or (a + b + c ) = 0

(a² + b² + c² -ab - ac -bc) = 0 cannot be zero because:

2a² + 2b² + 2c² -2ab - 2ac - 2bc = 0  

a² + b² -2ab + a² +b² +2c² - 2ac -2bc = 0  

(a-b)² + a² + c² -2ac + b² + c² -2bc = 0  

(a-b)² + (a-c)² + (b-c)² = 0

As a ≠ b ≠ c SO ,,, a + b + c = 0

2)

The value of x³-y³-z³-3xyz=0

 

Given x = y + z ----(1)

On Cubing equation (1) , we get

x³=(y + z)³

 we know the algebraic identity,

$\boxed {(a+b)^{3}=a^{3}+b^{3}+3ab(a+b)}$

=> x³=y³+z³+3yz(y + z)

=> x³=y³+z³+3yzx              /* From(1)*/

=> x³-y³-z³-3xyz = 0

3)

Given : x + y = 2 ;  $\frac{1}{x} + \frac{1}{y} = 2$

$\frac{x+y}{xy} = 2\\=> \frac{2}{xy} =2$

=> x × y = 1

$x^{3} + y^{3} = (x+y)^{3} - 3xy(x+y)$

$x^{3} +y^{3} = 2^{3} -3(1)(2)\\=> x^{3} +y^{3} = 8-6\\=> x^{3} +y^{3} = 2$

Hope it helps

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