can someone please help me simplify these three expressions if possible:(

...and if possible please show working out so i can understand better thanks

(a) (x+1)^3 -3x (x+1)

(b) y(2y+3) +2y^2 (y+1)(y-1)

(c) 14x^3 (3-2x) / x^2(x-1)(4x-6)

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- Sep 29th 2007, 05:39 AMraza88sabsimplifying expressions
can someone please help me simplify these three expressions if possible:(

...and if possible please show working out so i can understand better thanks

(a) (x+1)^3 -3x (x+1)

(b) y(2y+3) +2y^2 (y+1)(y-1)

(c) 14x^3 (3-2x) / x^2(x-1)(4x-6) - Sep 29th 2007, 05:46 AMtopsquark
This is similar in concept to

$\displaystyle a^3 - ba = a(a^2 - b)$

Here $\displaystyle a = x + 1$ and $\displaystyle b = 3x$

So

$\displaystyle (x + 1)^3 - 3x(x + 1) = (x + 1) ( (x + 1)^2 - 3x)$

$\displaystyle = (x + 1)(x^2 + 2x + 1 - 3x) = (x + 1)(x^2 -x + 1)$

-Dan - Sep 29th 2007, 05:48 AMtopsquark
- Sep 29th 2007, 05:52 AMtopsquark
$\displaystyle \frac{14x^3(3 - 2x)}{x^2(x - 1)(4x - 6)}$

Note that $\displaystyle 4x - 6 = 2(2x - 3) = -2(3 - 2x)$

So

$\displaystyle \frac{14x^3(3 - 2x)}{x^2(x - 1)(4x - 6)} = -\frac{14x^3(3 - 2x)}{2x^2(x - 1)(3 - 2x)}$

Now we may cancel the common $\displaystyle 3 - 2x$ and the common 2x^2, provided we remember that $\displaystyle x \neq 0, \frac{3}{2}$.

$\displaystyle \frac{14x^3(3 - 2x)}{x^2(x - 1)(4x - 6)} = -\frac{14x^3(3 - 2x)}{2x^2(x - 1)(3 - 2x)} = -\frac{7x}{x - 1};~x \neq 0, \frac{3}{2}$

-Dan - Sep 29th 2007, 05:58 AMraza88sabthanks
thanks dan mate....im just going to see if i can do this my self and lets see if it sticks in my head somehere lol.....much appriciated:)