Question

factorise a^3-125-2a+10​

Answer 1

The given expression is:

$a^3$ - 125 - 2a + 10​

We have to find the factorisation of $a^3$ - 125 - 2a + 10​.

Solution:

∴ $a^3$ - 125 - 2a + 10​

= $a^3$ - $5^3$ - 2a + 10​

Using the algebraic identity:

$a^3$ - $b^3$ = (a - b)( $a^2$ + ab + $b^2$)

= (a - 5)( $a^2$ + 5a + $5^2$) - 2a + 10​

= (a - 5)( $a^2$ + 5a + $25$) - 2(a - 5)

Taking (a - 5) as common, we get

= (a - 5)[( $a^2$ + 5a + $25$) - 2]

= (a - 5)( $a^2$ + 5a + $25$ - 2)

= (a - 5)( $a^2$ + 5a + 23)

∴ The factorisation of $a^3$ - 125 - 2a + 10 = (a - 5)( $a^2$ + 5a + 23)

Thus, the factorisation of $a^3$ - 125 - 2a + 10 is "(a - 5)( $a^2$ + 5a + 23)".

Answer 2

$\textbf{To find:}$

$\text{Factors of a^3-125-2a+10 }$

$\text{Consider,}$

$a^3-125-2a+10$

$=(a^3-125)-2(a-5)$

$=(a^3-5^3)-2(a-5)$

$\text{using the identity}$

$\boxed{\bf\,a^3-b^3=(a-b)(a^2+ab+b^2)}$

$=(a-5)(a^2+5a+25)-2(a-5)$

$=(a-5)((a^2+5a+25)-2)$

$=(a-5)(a^2+5a+23)$

$\therefore\boxed{\bf\,a^3-125-2a+10=(a-5)(a^2+5a+23)}$

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