Question

1/a^2-3a+2 + 1/a^2-5a+6 + 2/a^2-8a+15

Answer 1

Answer:

$\frac{4+16a^3+23a^3}{a^2}$

Step-by-step explanation:

Given

$1/a^2-3a+2 + 1/a^2-5a+6 + 2/a^2-8a+15$

Required

Simplify

$1/a^2-3a+2 + 1/a^2-5a+6 + 2/a^2-8a+15$

Collect Like Terms

$1/a^2 + 1/a^2 + 2/a^2-3a-5a-8a+2+6+15$

$1/a^2 + 1/a^2 + 2/a^2-16a+23$

Add Fractions:

$\frac{1+1+2}{a^2} + 16a + 23$

$\frac{4}{a^2} + 16a + 23$

This can be further expressed as:

$\frac{4+16a^3+23a^3}{a^2}$