Question

evaluate the following without finding the actual cubes:
21^3-20^3

plz answer.​

Answer 1

Answer:

Solution:

(i) (-12)³ + (7)³+ (5)³

Let x = -12, y = 7 & z = 5

x + y + z = -12 + 7 + 5 = 0

Here, x + y + z = 0

We know that if,

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

(-12)³ + (7)³ + (5)³ = 3(-12)(7)(5) = -1260

(-12)³+ (7)³+ (5)³ = - 1260

(ii) (28)³ + (–15)³ + (-13)³

Let x = 28, y = -15 and z = -13

x + y + z = 28– 15 – 13 = 0

Here, x + y + z = 0

We know that if, x + y + z = 0,

then x³ + y³ + z³ = 3xyz

(28)³+ (–15)³ + (-13)³ = 3(28)(-15)(-13) = 16380

(28)³+ (–15)³ + (-13)³ = 16380

Step-by-step explanation:

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