Question

Simplify the following
3 a²(a²– 1) +¼ a2(a+a) ¾a – (a³ - 1)​

Answer 1

Answer:

$\frac{12a^4-a^3-12a^2+4}{4}$

Step-by-step explanation:

$3a^{2} (a^{2} -1)+\frac{1}{4} a2(a+a)\frac{3}{4} a-(a^3-1)$

After expanding

$3a^4-3a^2+\frac{12}{16} a^3-a^3+1$

$3a^4-3a^2+\frac{3}{4} a^3-a^3+1$

$3a^4-3a^2+(\frac{3a^3-4a^3}{4})+1$

$3a^4-3a^2-\frac{a^3}{4} +1$

After taking the L.C.M

$\frac{12a^4-12a^2-a^3+4}{4}$

after rearranging we get,

$\frac{12a^4-a^3-12a^2+4}{4}$

Answer 2

Given:

$\bold{3a^2(a^2- 1) +\frac{1}{4} a^2(a+a)\frac{3}{4}a- (a^3 - 1)}$

To find:

Simplify

Solution:

$\Rightarrow 3a^2(a^2- 1) +\frac{1}{4} a^2(a+a)\frac{3}{4}a- (a^3 - 1)\\\\\Rightarrow 3a^4- 3a^2+\frac{1}{4} a^2(2a)\frac{3}{4}a- (a^3 - 1)\\\\\Rightarrow 3a^4- 3a^2+\frac{1}{4}\times2a^4\times\frac{3}{4}- (a^3 - 1)\\\\\Rightarrow 3a^4- 3a^2+\frac{6a^4}{16} - a^3 + 1\\\\\Rightarrow \frac{48a^4- 48a^2+6a^4-16a^3+16}{16} \\\\\Rightarrow \frac{54a^4-16a^3- 48a^2+16}{16} \\\\\Rightarrow \frac{2(27a^4-8a^3- 24a^2+8)}{16} \\\\\Rightarrow \frac{(27a^4-8a^3- 24a^2+8)}{8}$

The final answer is= $\bold{\frac{(27a^4-8a^3- 24a^2+8)}{8} }$