Ta có:
$a+b=a²+b²=a³+b³$
$⇒a+b+a³+b²=2.(a²+b²)$
$⇒(a−2a²+a³)+(b−2b²+b³)=0$
$⇒a.(1−2a+a²)+b.(1−2b+b²)=0$
$⇒a.(1−a)²+b.(1−b)²=0(1)$
Ta có:
$(1−a)²≥0$
$⇒a.(1−a)²≥0$
$(1−b)²≥0$
$⇒b.(1−b)²≥0$
Từ $(1)$ ta có:
$⇔$\(\left[ \begin{array}{l}a.(1−a)²=0\\b.(1−b)²=0\end{array} \right.\)
$⇔$\(\left[ \begin{array}{l}1-a=0\\1-b=0\end{array} \right.\)
$⇔a=b=1$
Vậy giá trị của $P$ là:
$P=a^{2015} +b^{2015}$
$P=1+1$
$P=2$
