I have been doing pascal triangle and do not know where this question comes from? solve for b

(3+b)to the 4th power = 1
do I take 3 to the 4th + b to the 4th =1 and then subtract 81 so that -81 =b to the fourth which = 3

1 answer

The 4th row of Pascal's Triangle is

1 4 6 4 1

That is,

(a+b)^4 = a^4 + 4a^3b + 6a^2b^2 + 4ab^3 + b^4

So, (3+b)^4 is

3^4 + 4(3^3)b + 6(3^2)b^2 + 4(3^1)b^3 + b^4
= 81 + 108b + 54b^2 + 12b^3 + b^4

That's all very nice, but if your question is to solve for b in

(3+b)^4 = 1

Consider the equation

x^4 = 1
x = ±1,±i
So, you have

3+b = ±1,±i
b = -3±1, -3±i
b = -4,-2,-3+i, -3-i
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