Question

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Answer 1

As we know from the formula,

A3 + B3 + C3 = ( A + B + C) (A2 + B2 + C2 – AB – BC – CA) + 3ABC

If A + B + C = 0 then,

A3 + B3 + C3 = 0 x (A2 + B2 + B2 – AB – BC – CA) + 3ABC

A3 + B3 + C3 = 0 + 3ABC

A3 + B3 + C3 - 3ABC= 0

As we know that from given question,

A = a, B = b and C = 1, Put these value in above equation, we will get

a3 + b3 + 13 - 3 x a x b x 1 = 0

a3 + b3 + 1 - 3ab = 0

 

4. Correct Option: C

x3 - x2y - xy2 + y3 = x3 + y3 - x2y - xy2

= (x + y)3 - 3xy(x + y) – xy(x + y)

= (x + y)3 - 4xy(x + y) = a3 - 4b2a

5. Correct option : C

Given,

x2+y2−4x−4y+8=0

or, x2−4x+4+y2−4y+4=0

or, (x−2)2+(y−2)2=0

This gives x−2=0,y−2=0⇒x=2,y=2.

x-y

2-2 =0

6. Correct Option: D

(a – 1)2 + (b + 2)2 + (c + 1)2 = 0

⇒  a – 1 = 0 ⇒ a = 1;

b + 2 = 0 ⇒ b = –2

c + 1 = 0 ⇒ c = –1

∴  2a – 3b + 7c

= 2 – 3 (– 2) + 7 (–1)

= 2 + 6 – 7 = 1

8. Correct Option: A

(3a + 1)2 + (b – 1)2 + (2c–3)2 = 0

⇒  3a + 1 = 0

⇒  3a = –1

b – 1 = 0

⇒  b = 1

2c – 3 = 0

⇒  2c = 3

∴  3a + b + 2c = –1 + 1 + 3 = 3

9. Correct option : A

(x+y)^x/y

(12+4)¹²/⁴

(16)³

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