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

Solution:

i) (-12)³ + (7)³ + (5)³

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

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

We know the algebraic identity, if x + y + z = 0 then x³ + y³ + z³ = 3xyz

Thus, (-12)³ + (7)³ + (5)³ = 3(-12)(7)(5)

= -1260

ii) (28)³ + (-15)³ + (-13)³

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

Then x + y + z = 28 - 15 - 13 = 0

We know the algebraic identity, if x + y + z = 0 then x³ + y³ + z³ = 3xyz

Thus, (28)³ + (-15)³ + (-13)³ = 3(28)(-15)(-13)

= 16380

☛ Check: Class 9 NCERT Solutions Maths Chapter 2

Video Solution:

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

NCERT Solutions Class 9 Maths Chapter 2 Exercise 2.5 Question 14

Summary:

The value of each of the following (−12)³ + (7)³ + (5)³ and (28)³ + (−15)³ + (−13)³ without actually calculating the cubes are -1260 and 16380 respectively.

☛ Related Questions:

• Give possible expressions for the length and breadth of each of the following rectangles, in which their areas are given:i) Area: 25a² - 35a + 12ii) Area: 35y² + 13y - 12
• What are the possible expressions for the dimensions of the cuboids whose volume are given below?i) Volume: 3x² - 12xii) Volume: 12ky² + 8ky - 20k
• Verify that x³ + y³ + z³ - 3xy = 1/2 (x + y + z)[(x - y)² + (y - z)² + (z - x)²]
• If x + y + z = 0, show that x³ + y³ + z³ = 3xyz.