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`



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