Algebra.

[MiedzianyDawid]

Wykaż, że jeśli \(x \neq 0\) i a-b=x, \(a^2-b^2=y\) i \(a^3-b^3=z\), to \(z= \frac{x^4+3y^2}{4x}\).

[kerajs]

\(P= \frac{x^4+3y^2}{4x}= \frac{1}{4}(x^3+3 \frac{y^2}{x} )= \frac{1}{4}((a-b)^3+3 \frac{(a^2-b^2)^2}{a-b} )= \frac{1}{4}(a^3-3a^2b+3ab^2-b^3+3(a^2-b^2)(a+b))=\\= \frac{1}{4}(a^3-3a^2b+3ab^2-b^3+3a^3+3a^2b-3ab^2-3b^3)=\frac{1}{4}(4a^3-4b^3) =a^3-b^3=z=L\)