Factorise-x 3-125A- x -5 x 2-5 x -25B- x -3 x 2-5 x -25- x -5 x 2-5 x -25D- x -5 x 2-10 x -25

The correct option is A (x + 5)(x^2 − 5x + 25)We have x^3 +125, 125 = 5×5×5 This can be written as x^3 + 5^3 Comparing x^3 + 5^3 with a^3 + b^3 we get a = x, b ...

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Byju's Answer

Standard IX

Mathematics

(x³±y³)

Factorise:x 3...

Question

Factorise:
$x^{3} + 125$

(x + 5) (x^{2} - 5 x + 25)

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(x + 3) (x^{2} - 5 x + 25)

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(x + 5) (x^{2} - 5 x + 25)

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(x + 5) (x^{2} - 10 x + 25)

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Solution

The correct option is A $(x + 5) (x^{2} - 5 x + 25)$
We have $x^{3} + 125$ ,
$125 = 5 \times 5 \times 5$
This can be written as $x^{3} + 5^{3}$
Comparing $x^{3} + 5^{3}$ with $a^{3} + b^{3}$
we get a = x, b = 5
Using identity,
$a^{3} + b^{3} = (a + b) (a^{2} - a b + b^{2})$
$x^{3} + 5^{3} = (x + 5) (x^{2} - (x) (5) + 5^{2})$
$x^{3} + 5^{3} = (x + 5) (x^{2} - 5 x + 25)$

Suggest Corrections

Factorise: x3+125

The correct option is A (x+5)(x2−5x+25)We have x3+125, 125=5×5×5 This can be written as x3+53 Comparing x3+53 with a3+b3 we get a = x, b = 5 Using identity, a3+b3=(a+b)(a2−ab+b2) x3+53=(x+5)(x2−(x)(5)+52) x3+53=(x+5)(x2−5x+25)

Factorise:
$x^{3} + 125$