Question

Without actually calculating cubes, find the value of each of the following: (iv) (28)^3 + (−15)^3 + (−13)^3

Answer 1

Answer:

(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:

Answer 2

Answer:

$\huge\bold {★} \green{A} \blue{N} \red{S} \purple{W} \orange{E}{R}$

★ From the above question,we have x^3+y^3+z^3=3xyz, If x+y+z=0

→ Here 28+(-15)+(-13)=0

So,(28)^3+(-15)^3+(-13)^3

=3×28(-15)(-13)=16380