Enter a problem...
Basic Math Examples
a1-a2+a1+a2
Step 1
Step 1.1
Rewrite 1 as 12.
a12-a2+a1+a2
Step 1.2
Since both terms are perfect squares, factor using the difference of squares formula, a2-b2=(a+b)(a-b) where a=1 and b=a.
a(1+a)(1-a)+a1+a2
a(1+a)(1-a)+a1+a2
Step 2
To write a(1+a)(1-a) as a fraction with a common denominator, multiply by 1+a21+a2.
a(1+a)(1-a)⋅1+a21+a2+a1+a2
Step 3
To write a1+a2 as a fraction with a common denominator, multiply by (1+a)(1-a)(1+a)(1-a).
a(1+a)(1-a)⋅1+a21+a2+a1+a2⋅(1+a)(1-a)(1+a)(1-a)
Step 4
Step 4.1
Multiply a(1+a)(1-a) by 1+a21+a2.
a(1+a2)(1+a)(1-a)(1+a2)+a1+a2⋅(1+a)(1-a)(1+a)(1-a)
Step 4.2
Multiply a1+a2 by (1+a)(1-a)(1+a)(1-a).
a(1+a2)(1+a)(1-a)(1+a2)+a((1+a)(1-a))(1+a2)((1+a)(1-a))
Step 4.3
Reorder the factors of (1+a)(1-a)(1+a2).
a(1+a2)(1+a)(1+a2)(1-a)+a((1+a)(1-a))(1+a2)((1+a)(1-a))
Step 4.4
Reorder the factors of (1+a2)((1+a)(1-a)).
a(1+a2)(1+a)(1+a2)(1-a)+a((1+a)(1-a))(1+a)(1+a2)(1-a)
a(1+a2)(1+a)(1+a2)(1-a)+a((1+a)(1-a))(1+a)(1+a2)(1-a)
Step 5
Combine the numerators over the common denominator.
a(1+a2)+a((1+a)(1-a))(1+a)(1+a2)(1-a)
Step 6
Step 6.1
Factor a out of a(1+a2)+a((1+a)(1-a)).
a(1+a2+(1+a)(1-a))(1+a)(1+a2)(1-a)
Step 6.2
Let u=1. Substitute u for all occurrences of 1.
Step 6.2.1
Subtract a2 from a2.
a(u+u2+0)(1+a)(1+a2)(1-a)
Step 6.2.2
Add u+u2 and 0.
a(u+u2)(1+a)(1+a2)(1-a)
a(u+u2)(1+a)(1+a2)(1-a)
Step 6.3
Factor u out of u+u2.
Step 6.3.1
Raise u to the power of 1.
a(u1+u2)(1+a)(1+a2)(1-a)
Step 6.3.2
Factor u out of u1.
a(u⋅1+u2)(1+a)(1+a2)(1-a)
Step 6.3.3
Factor u out of u2.
a(u⋅1+u⋅u)(1+a)(1+a2)(1-a)
Step 6.3.4
Factor u out of u⋅1+u⋅u.
a(u(1+u))(1+a)(1+a2)(1-a)
a(u(1+u))(1+a)(1+a2)(1-a)
Step 6.4
Replace all occurrences of u with 1.
a(1(1+1))(1+a)(1+a2)(1-a)
Step 6.5
Add 1 and 1.
a(1⋅2)(1+a)(1+a2)(1-a)
Step 6.6
Multiply 2 by 1.
a⋅2(1+a)(1+a2)(1-a)
a⋅2(1+a)(1+a2)(1-a)
Step 7
Move 2 to the left of a.
2a(1+a)(1+a2)(1-a)