Question

Please answer the following 2 qs​

Answer 1

Questions:-

1. If x + y + z = 0 , show that x³ + y³ + z³ = 3xyz.
2. Without actually calculating the cubes , find the value of each of the following:

i) ( - 12)³ + (7)³ + (5)³

ii) (28)³ + ( - 15)³ + ( - 13)³

Answers:-

1. Given:-

x + y + z = 0 -- equation (1)

We know that,

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

So,

⟹ x³ + y³ + z³ - 3xyz = (x + y + z)(x² + y² + z² - xy - yz - zx)

⟹ x³ + y³ + z³ - 3xyz = 0 * (x² + y² + z² - xy - yz - zx)

[ ∵ From equation (1) ]

⟹ x³ + y³ + z³ - 3xyz = 0

⟹ x³ + y³ + z³ = 3xyz

Hence, Proved.

_________________________

2. (i) ( - 12)³ + (7)³ + (5)³

We know,

If x + y + z = 0 , then x³ + y³ + z³ = 3xyz.

Let,

• x = - 12
• y = 7
• z = 5

⟹ x + y + z = - 12 + 7 + 5 = 0

Therefore,

⟹ ( - 12)³ + (7)³ + (5)³ = 3( - 12)(7)(5)

⟹ ( - 12)³ + (7)³ + (5)³ = - 1260

ii) (28)³ + ( - 15)³ + ( - 13)³

In the same way;

⟹ (28)³ + ( - 15)³ + ( - 13)³ = 3(28)(- 15)(- 13)

⟹ (28)³ + ( - 15)³ + ( - 13)³ = 16380

Answer 2

Step-by-step explanation:

13)

Given :-

x + y + z = 0

To Show :-

x³ + y³ + z³ = 3xyz

Solution :-

x + y + z = 0 [Given]

x³ + y³ + z³ = 3xyz [To Prove]

On transposing 3xyz to LHS

x³ + y³ + z³ - 3xyz = (x + y + z)[x² + y² + z² - (xy + yz + xz)]

x³ + y³ + z³ - 3xyz = 0[x² + y² + z² - (xy + yz + xz)]

x³ + y³ + z³ - 3xyz = 0[x² + y² + z² - xy - yz - xz]

x³ + y³ + z³ - 3xyz = 0

On transposing 3xyz to RHS

x³ + y³ + z³ = 0 + 3xyz

x³ + y³ + z³ = 3xyz

Hence proved

14)

(i) (-12)³ - (7)³ + (5)³

From above question we get

If x + y + z = 0 then x³ + y³ + z³ = 3xyz

x + y + z = -12 + 7 + 5

x + y + z = -12 + 12

x + y + z = 0

So,

x³ + y³ + z³ = 3(-12)(7)(5)

x³ + y³ + z³ = -36 × 35

x³ + y³ + z³ = -126/

(ii) (28)³ + (-15)³ + (-13)³

Same from above

x + y + z = 28 + (-15) + (-13)

x + y + z = 28 - 15 - 13

x + y + z = 28 - 28

x + y + z = 0

So,

x³ + y³ + z³ = 3(28)(-15)(-13)

x³ + y³ + z³ = 72 × 195

x³ + y³ + z³ = 16380