Algebra Examples

Expand Using Pascal's Triangle (a+b)^6
(a+b)6
Step 1
Pascal's Triangle can be displayed as such:
1
1-1
1-2-1
1-3-3-1
1-4-6-4-1
1-5-10-10-5-1
1-6-15-20-15-6-1
The triangle can be used to calculate the coefficients of the expansion of (a+b)n by taking the exponent n and adding 1. The coefficients will correspond with line n+1 of the triangle. For (a+b)6, n=6 so the coefficients of the expansion will correspond with line 7.
Step 2
The expansion follows the rule (a+b)n=c0anb0+c1an-1b1+cn-1a1bn-1+cna0bn. The values of the coefficients, from the triangle, are 1-6-15-20-15-6-1.
1a6b0+6a5b+15a4b2+20a3b3+15a2b4+6ab5+1a0b6
Step 3
Substitute the actual values of a a and b b into the expression.
1(a)6(b)0+6(a)5(b)1+15(a)4(b)2+20(a)3(b)3+15(a)2(b)4+6(a)1(b)5+1(a)0(b)6
Step 4
Simplify each term.
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Step 4.1
Multiply (a)6 by 1.
(a)6(b)0+6(a)5(b)1+15(a)4(b)2+20(a)3(b)3+15(a)2(b)4+6(a)1(b)5+1(a)0(b)6
Step 4.2
Anything raised to 0 is 1.
a61+6(a)5(b)1+15(a)4(b)2+20(a)3(b)3+15(a)2(b)4+6(a)1(b)5+1(a)0(b)6
Step 4.3
Multiply a6 by 1.
a6+6(a)5(b)1+15(a)4(b)2+20(a)3(b)3+15(a)2(b)4+6(a)1(b)5+1(a)0(b)6
Step 4.4
Simplify.
a6+6a5b+15(a)4(b)2+20(a)3(b)3+15(a)2(b)4+6(a)1(b)5+1(a)0(b)6
Step 4.5
Simplify.
a6+6a5b+15a4b2+20a3b3+15a2b4+6a(b)5+1(a)0(b)6
Step 4.6
Multiply (a)0 by 1.
a6+6a5b+15a4b2+20a3b3+15a2b4+6ab5+(a)0(b)6
Step 4.7
Anything raised to 0 is 1.
a6+6a5b+15a4b2+20a3b3+15a2b4+6ab5+1(b)6
Step 4.8
Multiply (b)6 by 1.
a6+6a5b+15a4b2+20a3b3+15a2b4+6ab5+b6
a6+6a5b+15a4b2+20a3b3+15a2b4+6ab5+b6
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