Answer: 24*sqrt(2)
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Work Shown:
A = 1+sqrt(2)
B = 1-sqrt(2)
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Using A and B, let's add and subtract those values
A+B = (1+sqrt(2))+(1-sqrt(2))
A+B = 1+sqrt(2)+1-sqrt(2)
A+B = 1+1+sqrt(2)-sqrt(2)
A+B = 2
and
A-B = (1+sqrt(2))-(1-sqrt(2))
A-B = 1+sqrt(2)-1+sqrt(2)
A-B = 1-1+sqrt(2)+sqrt(2)
A-B = 2sqrt(2)
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Next, square both sides of the A equation
A^2 = (1+sqrt(2))^2
A^2 = (1+sqrt(2))(1+sqrt(2))
A^2 = 1(1+sqrt(2))+sqrt(2)(1+sqrt(2))
A^2 = 1+sqrt(2)+sqrt(2)+sqrt(2)*sqrt(2)
A^2 = 1+2sqrt(2)+2
A^2 = 3+2sqrt(2)
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Do the same for B
B^2 = (1-sqrt(2))^2
B^2 = (1-sqrt(2))(1-sqrt(2))
B^2 = 1(1-sqrt(2))-sqrt(2)(1-sqrt(2))
B^2 = 1-sqrt(2)-sqrt(2)-sqrt(2)*(-sqrt(2))
B^2 = 1-2sqrt(2)+2
B^2 = 3-2sqrt(2)
There's a bit of nice symmetry going on with A^2 and B^2, due to A and B having similar symmetric structures themselves.
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Lets add A^2 to B^2
A^2+B^2 = (3+2sqrt(2)) + (3-2sqrt(2))
A^2+B^2 = 3+2sqrt(2) + 3-2sqrt(2)
A^2+B^2 = 3+3+2sqrt(2)-2sqrt(2)
A^2+B^2 = 6
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Now onto the final event
We'll use the difference of squares rule twice
Afterward, apply substitution using the previous sections above.
A^4 - B^4 = (A^2)^2 - (B^2)^2
A^4 - B^4 = (A^2-B^2)(A^2+B^2)
A^4 - B^4 = (A-B)(A+B)(A^2+B^2)
A^4 - B^4 = (2sqrt(2))(2)(6)
A^4 - B^4 = 2*6*2*sqrt(2)
A^4 - B^4 = 24*sqrt(2)
(1+sqrt(2))^4 - (1-sqrt(2))^4 = 24*sqrt(2)
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To check this with your calculator, you can type in the following
(1+sqrt(2))^4 - (1-sqrt(2))^4 - 24*sqrt(2)
The result you should get should be 0, or very close to it
I'm using the property that if x = y, then x-y = 0.