Examples

Expand Using the Binomial Theorem
(x-1)3(x1)3
Step 1
Use the binomial expansion theorem to find each term. The binomial theorem states (a+b)n=nk=0nCk(an-kbk)(a+b)n=nk=0nCk(ankbk).
3k=03!(3-k)!k!(x)3-k(-1)k3k=03!(3k)!k!(x)3k(1)k
Step 2
Expand the summation.
3!(3-0)!0!(x)3-0(-1)0+3!(3-1)!1!(x)3-1(-1)1+3!(3-2)!2!(x)3-2(-1)2+3!(3-3)!3!(x)3-3(-1)33!(30)!0!(x)30(1)0+3!(31)!1!(x)31(1)1+3!(32)!2!(x)32(1)2+3!(33)!3!(x)33(1)3
Step 3
Simplify the exponents for each term of the expansion.
1(x)3(-1)0+3(x)2(-1)1+3(x)1(-1)2+1(x)0(-1)31(x)3(1)0+3(x)2(1)1+3(x)1(1)2+1(x)0(1)3
Step 4
Simplify each term.
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Step 4.1
Multiply (x)3(x)3 by 11.
(x)3(-1)0+3(x)2(-1)1+3(x)1(-1)2+1(x)0(-1)3(x)3(1)0+3(x)2(1)1+3(x)1(1)2+1(x)0(1)3
Step 4.2
Anything raised to 00 is 11.
x31+3(x)2(-1)1+3(x)1(-1)2+1(x)0(-1)3x31+3(x)2(1)1+3(x)1(1)2+1(x)0(1)3
Step 4.3
Multiply x3x3 by 11.
x3+3(x)2(-1)1+3(x)1(-1)2+1(x)0(-1)3x3+3(x)2(1)1+3(x)1(1)2+1(x)0(1)3
Step 4.4
Evaluate the exponent.
x3+3x2-1+3(x)1(-1)2+1(x)0(-1)3x3+3x21+3(x)1(1)2+1(x)0(1)3
Step 4.5
Multiply -11 by 33.
x3-3x2+3(x)1(-1)2+1(x)0(-1)3
Step 4.6
Simplify.
x3-3x2+3x(-1)2+1(x)0(-1)3
Step 4.7
Raise -1 to the power of 2.
x3-3x2+3x1+1(x)0(-1)3
Step 4.8
Multiply 3 by 1.
x3-3x2+3x+1(x)0(-1)3
Step 4.9
Multiply (x)0 by 1.
x3-3x2+3x+(x)0(-1)3
Step 4.10
Anything raised to 0 is 1.
x3-3x2+3x+1(-1)3
Step 4.11
Multiply (-1)3 by 1.
x3-3x2+3x+(-1)3
Step 4.12
Raise -1 to the power of 3.
x3-3x2+3x-1
x3-3x2+3x-1
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