Factorize a5 b-a b5A- a ba2-b2B- a-ba-bC- a ba2-b2a-ba-bD- a-ba2-b2

Question

Factorize $a^{5} b - a b^{5}$ .
$a b (a^{2} + b^{2})$

$(a + b) (a - b)$

$a b (a^{2} + b^{2}) (a + b) (a - b)$
$(a - b) (a^{2} + b^{2})$

Solution

The correct option is C $a b (a^{2} + b^{2}) (a + b) (a - b)$
$a b (a^{2} + b^{2}) (a + b) (a - b)$

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Similar questions

Mathematics | RD Sharma Standard IX

Q1.

$The product (a + b) (a - b) (a^{2} - a b + b^{2}) (a^{2} + a b + b^{2})$ is equal to

Mathematics

Q2.

x = {(\frac{a}{b})}^{\frac{2 a b}{a^{2} - b^{2}}}

, shew that

\frac{a b}{a^{2} + b^{2}} ⎛ ⎜ ⎝ x^{\frac{a}{b}} + x^{\frac{b}{a}} ⎞ ⎟ ⎠ = {(\frac{a}{b})}^{\frac{a^{2} + b^{2}}{a^{2} - b^{2}}}

Maths

Q3.

Show that the determinant

∣ ∣ ∣ ∣ ∣ \begin{matrix} a^{2} + b^{2} + c^{2} & b c + c a + a b & b c + c a + a b b c + c a + a b & a^{2} + b^{2} + c^{2} & b c + c a + a b b c + c a + a b & b c + c a + a b & a^{2} + b^{2} + c^{2} \end{matrix} ∣ ∣ ∣ ∣ ∣

is always non-negative.

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Factorize $a^{5} b - a b^{5}$ .
$a b (a^{2} + b^{2})$

$(a + b) (a - b)$

$a b (a^{2} + b^{2}) (a + b) (a - b)$
$(a - b) (a^{2} + b^{2})$

The correct option is C $a b (a^{2} + b^{2}) (a + b) (a - b)$
$a b (a^{2} + b^{2}) (a + b) (a - b)$

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Factorize a5b−ab5.ab(a2+b2) (a+b)(a−b) ab(a2+b2)(a+b)(a−b)(a−b)(a2+b2)

The correct option is C ab(a2+b2)(a+b)(a−b)ab(a2+b2)(a+b)(a−b)

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Factorize $a^{5} b - a b^{5}$ .
$a b (a^{2} + b^{2})$

$(a + b) (a - b)$

$a b (a^{2} + b^{2}) (a + b) (a - b)$
$(a - b) (a^{2} + b^{2})$

The correct option is C $a b (a^{2} + b^{2}) (a + b) (a - b)$
$a b (a^{2} + b^{2}) (a + b) (a - b)$