foloseste-te de formula (a-b)(a+b) = a^2-b^2
(2x - 3)(2x + 3) = 4x^2-9
(√2 x + 1)(√2 x -1) = 2x-1
(4a-8)(a+2)(a²+4)=4(a-2)(a+2)(a²+4)=4(a²-4)(a²+4)=4(a^4-16)=4a^4-64
(3a²-2)(3x²+2) - aici sigur nu e aceeasi necunoscuta in ambele paranteze? Daca da => (3x²-2)(3x²+2)=9x^4-4
(4x + 3y)(4x-3y)=16x^2-9y^2
(3x²-1)(3x²+1)=9x^4-1
(3x² -2√3)(3x² +2√3) = 9x^4-12
(a-1)(a+1)(a²+1)(a⁴+1)=(a²-1)(a²+1)(a⁴+1)=(a⁴-1)(a⁴+1)=a^8-1
(x²+1+√2)(x²+1-√2)=(x²+1)^2-2=x^4+2x^2+1-2=x^4+2x^2-1
(1/2 x - √2)(1/2 x+√2)=1/4x^2-2
(x-2)(x+2)(x²+4)(x⁴+16)=(x²-4)(x²+4)(x⁴+16)=(x⁴-16)(x⁴+16)=x^8-256
P.S. x^y inseamna x la puterea y