Bài 1:
Ta có: \(a^3+b^3+c^3=\left(a+b+c\right)^3-3\left(a+b\right)\left(a+c\right)\left(b+c\right)\)(*)
Ta có: a+b+c=0
\(\Leftrightarrow a=0-b-c\)
Thay a+b+c=0 và a=0-b-c vào biểu thức (*), ta được:
\(a^3+b^3+c^3=0^3-\left(0-b-c+b\right)\left(0-b-c+c\right)\left(b+c\right)\)
\(=0-\left(0-c\right)\cdot\left(0-b\right)\cdot\left(b+c\right)\)
\(=0-\left(-c\right)\cdot\left(-b\right)\cdot\left(b+c\right)\)
\(=0-bc\left(b+c\right)\)
\(=-bc\left(b+c\right)\)(1)
Ta có: a+b+c=0
\(\Leftrightarrow b+c=-a\)
Thay b+c=-a vào biểu thức (1), ta được:
\(a^3+b^3+c^3=-bc\cdot\left(-a\right)=abc\)(đpcm)
Bài 2:
Ta có: \(\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2+1\right)\left(2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)\)
\(=\left(2^{16}-1\right)\left(2^{16}+1\right)\)
\(=2^{32}-1\)
