# Choose one of the factors of x^{6} + 1000.

x^{2}, x^{2} - 10, x^{4} - 10x^{2} + 100, x^{4} + 10x^{2} + 100

**Solution:**

Let us use (a + b)^{3} identity to factorize the expression. (a + b)^{3 }= a^{3} + b^{3} + 3ab^{2} + 3a^{2}b can be rewritten as

⇒ a^{3} + b^{3} = (a + b)^{3} - 3ab^{2} - 3a^{2}b --- (1)

Therefore we can write the given expression x^{6} + 1000 as per equation(1)

x^{6} + 1000 = (x^{2})^{3} + (10)^{3} = (x^{2} + 10)^{3} - 3(x^{2})(10)^{2} - 3(x^{2})^{2}(10) --- (2)

Where x^{2} corresponds to “a” in equation(1) and 10 corresponds to “b” in equation(1)

Rewriting equation(2) we have,

(x^{2} + 10)^{3} - 300x^{2} - 30x^{4}

⇒ (x^{2} + 10)^{3} - 30x^{2}(x^{2} + 10) --- (3)

Taking (x^{2} + 10) as the common factor we have

(x^{2} + 10){(x^{2} + 10)^{2} - 30x^{2}}

⇒ (x^{2} + 10)(x^{4} + 100 + 20x^{2} - 30x^{2})

⇒ (x^{2 }+ 10)(x^{4} + 100 + 20x^{2} - 30x^{2})

⇒ (x^{2} + 10)(x^{4} - 10x^{2} + 100 )

The above expressions show x^{6} + 1000 has two factors and they are

(x^{2} + 10) & (x^{4 }- 10x^{2} + 100) and one of the two factors is (x^{4} - 10x^{2} + 100) which is the correct alternative of the choices given in the problem.

## Choose one of the factors of x^{6} + 1000.

**Summary:**

One of the other factor of x^{6} + 1000 is x^{4} - 10x^{2} + 100.

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