Class 9 Maths

# Polynomials

## Exercise 2.5 Part 11

Question 12: Verify that

x^3 + y^3 + z^3 - 3xyz

= ½(x + y + z)[(x – y)^2 + ( y – z)^2 + (z – x)^2]

Answer: RHS = ½(x + y + z)[(x – y)^2 + ( y – z)^2 + (z – x)^2]

Using the identity (a – b)^2 = a^2 + b^2 - 2ab

We get: ½(x + y + z)[(x^2 + y^2 - 2xy) + (y^2 + z^2 - 2yz) + (z^2 + x^2 - 2xz)]

= ½(x + y + z)(x^2 + y^2 - 2xy + y^2 + z^2 - 2yz + x^2 + z^2 - 2xz)

= ½(x + y + z)(x^2 + x^2 + y^2 + y^2 + z^2 - 2xy – 2yz – 2xz)

= ½(x + y + z).2(x^2 + y^2 + z^2 - xy – yz – xz)

= (x + y + z)(x^2 + y^2 + z^2 - xy – yz – xz

= x^3 + y^3 + z^3 - 3xyz = LHS

Question 13: If x + y + z = 0, show that x^3 + y^3 + z^3 = 3xyz

Answer: We know that, x^3 + y^3 + z^3 - 3xyz

= (x + y + z)(x^2 + y^2 + z^2 - xy – yz – xz)

After replacing (x + y + z) = 0 we get:

x^3 + y^3 + z^3 - 3xyz

Hence, we have:
(0)(x^2 + y^2 + z^2 – xy – yz – xz)

Or, x^3 + y^3 + z^3 - 3xyz = 0

Or, x^3 + y^3 + z^3 = 3xyz proved

Question 14: Without actually calculating the cubes, find the value of each of the following:

(i) ( - 12)^3 + 7^3 + 5^3

Answer: Given: ( - 12)^3 + 7^3 + 5^3

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

Now, x + y + z = – 12 + 7 + 5 = 0

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

Then, x^3 + y^3 + z^3 = 3xyz

Hence, (- 12)^3 + 7^3 + 5^3 = 3 xx ( - 12)xx7xx5

= 3 xx ( - 420) = - 1260

(ii) 28^3 + ( - 15)^3 + ( - 13)^3

Answer: Given; 28^3 + ( - 15)^3 + ( - 13)^3

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

Now, x + y + z = 28 – 15 – 13 = 0

Or, x + y + z = 0

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

Then, x^3 + y^3 + z^3 = 3xyz

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

= 3 xx 28 xx ( - 15) ( - 13)= 84 xx 195 = 16380