Without actually calculating the cubes, find the values of:
(i) (27)3 + (-17)3 + (-10)3
(ii) (-28)3 + (15)3 + (13)3
Answer :
(i) (27)3 + (-17)3 + (-10)3
Consider a = 27, b = – 17 and c = – 10
We know that
a + b + c = 27 – 17 – 10 = 0
So a + b + c = 0
a3 + b3 + c3 = 3abc
Substituting the values
273 + (-17)3 + (-10)3 = 3 (27) (-17) (- 10)
= 13770
(ii) (-28)3 + (15)3 + (13)3
Consider a = – 28, b = 15 and c = 13
We know that
a + b + c = – 28 + 15 + 13 = 0
So a + b + c = 0
a3 + b3 + c3 = 3abc
Substituting the values
(-28)3 + (15)3 + (13)3 = 3 (- 28) (15) (13)
= – 16380
More Solutions:
- If x + y = 6 and x – y = 4, find
- If x – 3 = 1/x, find the value of x2 + 1/x2.
- If x + y = 8 and xy = 3 ¾, find the values of
- Find the value of 2 (x + y)2 + (x – y)2.
- If a – b = 3 and ab = 4, find a3 – b3.
- Find the value of 8a3 – 27b3.
- If x + 1/x = 4, find the values of
- If x – 1/x = 5, find the value of x4 + 1/x4.
- If x – 1/x = √5, find the values of
- If x + 1/x = 6, find