calculate without using tables the values of
(1+√2)⁴ - (1-√2)⁴​

=========================================================

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)

-------------------

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.

Please help, due in 6 hrs (will mark brainliest and giving 85 points) will ask seperate question for the 85 points

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