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Expand (a+b)3 in descending powers of a

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Answer: a^3 + 3a^2b + 3ab^2 + b^3

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Step-by-step explanation:

The coefficients of each term, from left to right are 1, 3, 3, 1

These coefficient values are found in Pascal's triangle in the third row.

The variable terms will have variable terms in the form a^m*b^n. To start off, m = 3 and n = 0 to get a^3*b^0 = a^3*1 = a^3. From there you will have m count down (3,2,1,0) and n will count up (0,1,2,3)

So for the second term we have m = 2 and n = 1 to get the variable term a^2*b^1 = a^2b

The third term has m = 1 and n = 2 to get the variable term a^1*b^2 = ab^2

Finally the last term has m = 0 and n = 3 to get a^0*b^3 = 1*b^3 = b^3

Those variable terms are them multiplied with the coefficients mentioned previously to get the final answer.

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Another way you can do this problem is to use the distributive property and FOIL rule

(a+b)^3 = (a+b)(a+b)^2

(a+b)^3 = (a+b)(a^2+2ab+b^2) .... FOIL rule

(a+b)^3 = a(a^2+2ab+b^2) + b(a^2+2ab+b^2) ... distribute

(a+b)^3 = a*a^2+a*2ab+a*b^2 + b*a^2+b*2ab+b*b^2 ... distribute again

(a+b)^3 = a^3 + 2a^2b + ab^2 + a^2b + 2ab^2 + b^3

(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3

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