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Q14(ii):
Without actually calculating the cubes, find the value of each of the following: (ii) $(28)^3 + (–15)^3 + (–13)^3$

Solution :

Given Variables & Initial Setup

We are required to evaluate the algebraic expression $(28)^3 + (-15)^3 + (-13)^3$ without explicitly calculating the cubes of the individual numbers. To do this, we map the given terms to standard algebraic variables.

Let us define the base variables:

  • $a = 28$
  • $b = -15$
  • $c = -13$

Step 1: Evaluating the Sum of the Base Variables

Before applying any higher-degree polynomial identity, we must first test the linear sum of the defined variables, $a + b + c$.

Substituting the assigned values:

$a + b + c = 28 + (-15) + (-13)$

$a + b + c = 28 - 15 - 13$

$a + b + c = 28 - 28 = 0$

[By the fundamental properties of arithmetic addition, the sum of the base values is exactly zero.]

Step 2: Applying the Conditional Algebraic Identity

We utilize the standard polynomial identity for the sum of three cubes:

$a^3 + b^3 + c^3 - 3abc = (a + b + c)(a^2 + b^2 + c^2 - ab - bc - ca)$

[Per the Zero Product Property], if the first factor on the right-hand side $(a + b + c)$ is equal to $0$, the entire right-hand side of the equation evaluates to $0$.

Therefore, the identity simplifies to:

$a^3 + b^3 + c^3 - 3abc = 0 \implies a^3 + b^3 + c^3 = 3abc$

a = 28 b = -15, c = -13 Condition Met a + b + c = 0 Applied Identity a³ + b³ + c³ = 3abc

Step 3: Computing the Final Value

Since we have established that $a + b + c = 0$, we can directly substitute our values into the simplified identity $a^3 + b^3 + c^3 = 3abc$.

$(28)^3 + (-15)^3 + (-13)^3 = 3 \cdot (28) \cdot (-15) \cdot (-13)$

Now, we perform the sequential multiplication:

  • First, multiply the negative integers to determine the sign and partial product:
    $(-15) \times (-13) = +195$
    [The product of two negative numbers yields a positive scalar.]
  • Next, multiply the remaining constants:
    $3 \times 28 = 84$
  • Finally, multiply the two partial products together:
    $84 \times 195$

To compute $84 \times 195$ efficiently without a calculator, we can use the distributive property of multiplication over subtraction:

$84 \times (200 - 5) = (84 \times 200) - (84 \times 5)$

$= 16800 - 420$

$= 16380$

Final Solution: 16380


More Questions from Class 9 Mathematics Polynomials EXERCISE 2.4


CBSE Solutions for Class 9 Mathematics Polynomials


Chapters in CBSE - Class 9 Mathematics


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