Examples

Expand Using the Binomial Theorem
(x9)3
Step 1
Use the binomial expansion theorem to find each term. The binomial theorem states (a+b)n=nk=0nCk(ankbk).
3k=03!(3k)!k!(x)3k(9)k
Step 2
Expand the summation.
3!(30)!0!(x)30(9)0+3!(31)!1!(x)31(9)1+3!(32)!2!(x)32(9)2+3!(33)!3!(x)33(9)3
Step 3
Simplify the exponents for each term of the expansion.
1(x)3(9)0+3(x)2(9)1+3(x)1(9)2+1(x)0(9)3
Step 4
Simplify each term.
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Step 4.1
Multiply (x)3 by 1.
(x)3(9)0+3(x)2(9)1+3(x)1(9)2+1(x)0(9)3
Step 4.2
Anything raised to 0 is 1.
x31+3(x)2(9)1+3(x)1(9)2+1(x)0(9)3
Step 4.3
Multiply x3 by 1.
x3+3(x)2(9)1+3(x)1(9)2+1(x)0(9)3
Step 4.4
Evaluate the exponent.
x3+3x29+3(x)1(9)2+1(x)0(9)3
Step 4.5
Multiply 9 by 3.
x327x2+3(x)1(9)2+1(x)0(9)3
Step 4.6
Simplify.
x327x2+3x(9)2+1(x)0(9)3
Step 4.7
Raise 9 to the power of 2.
x327x2+3x81+1(x)0(9)3
Step 4.8
Multiply 81 by 3.
x327x2+243x+1(x)0(9)3
Step 4.9
Multiply (x)0 by 1.
x327x2+243x+(x)0(9)3
Step 4.10
Anything raised to 0 is 1.
x327x2+243x+1(9)3
Step 4.11
Multiply (9)3 by 1.
x327x2+243x+(9)3
Step 4.12
Raise 9 to the power of 3.
x327x2+243x729
x327x2+243x729
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