Question

13. If x+y+2=0, show that x +y +z = 3xyz.
Without actually calculating the cubes, find the value of each of the
(-12) + (7) + (5)
(28)3 + (-15) + (-13)
oh of the followir​

Answer 1

1st Answer :

Given :

x + y + z = 0

Required to Prove :

1. x + y + z = 3xyz

Proof :

Consider the first statement .

Using this first statement we are going to prove it .

Hence ,

x + y + z = 0

Now transpose z to the right side

x + y = -z

=> Let's do Cubing on both sides .

$(x + y {)}^{3} = (z {)}^{3}$

Here expand the left side using an identity

${x}^{3} + {y}^{3} + 3 {x}^{2} y + 3x {y}^{2} = { - z}^{3} \\ {x}^{3} + {y}^{3} + 3xy(x + y) = { - z}^{3} \\ {x}^{3} + {y}^{3} + 3xy( - z) = - {z}^{3} \\ {x}^{3} + {y}^{3} - 3xyz = { - z}^{3} \\ therefore \\ {x}^{3} + {y}^{3} + {z}^{3} = 3xyz$

Therefore;

It is proved .

2nd answer :

Given :

• (-12)^3+(7)^3+(5)^3
• (28)^3+(-15)^3+(-13)^3

Required to find:

1. Sum of the Cubes of the numbers.

Condition Mentioned:

• Without performing actual division

Solution:

Let consider an identity to solve this question

The identity is

${x}^{3} + {y}^{3} + {z}^{ 3} - 3xyz = 0$

Now transpose 3xyz to that side we get

${x}^{3} + {y}^{3} + {z}^{ 3} = 3xyz$

Now from the above it is proved that 3xyz will give us the sum of the cubes of the numbers

Hence ,

1. 3 × -12 × 7 × 5

= -1260

Therefore ,

The 1st bit answer is 1260 .

2. 3 × 28 × -15 × -13

= 16380

Therefore,

The 2nd bit answer is 16380