Question

đặt nhân tử chung (thêm bớt hạng tử)

a¹⁶+a⁸b⁸+b¹⁶

Answer 1

\(a^{16}+a^8b^8+b^{16}\)

\(=a^{16}+b^{16}+2a^8b^8-a^8b^8\)

\(=\left(a^8+b^8\right)^2-a^8b^8\)

\(=\left(a^8+b^8+a^4b^4\right)\left(a^8-a^4b^4+b^8\right)\)

\(=\left(a^8+b^8+2a^4b^4-a^4b^4\right)\left(a^8-a^4b^4+b^8\right)\)

\(=\left(\left(a^4+b^4\right)^2-a^4b^4\right)\left(a^8-a^4b^4+b^8\right)\)

\(=\left(a^4+b^4+a^2b^2\right)\left(a^4-a^2b^2+b^4\right)\left(a^8-a^4b^4+b^8\right)\)

\(=\left(a^4+b^4+2a^2b^2-a^2b^2\right)\left(a^4-a^2b^2+b^4\right)\left(a^8-a^4b^4+b^8\right)\)

\(=\left(\left(a^2+b^2\right)^2-a^2b^2\right)\left(a^4-a^2b^2+b^4\right)\left(a^8-a^4b^4+b^8\right)\)

\(=\left(a^2-ab+b^2\right)\left(a^2+ab+b^2\right)\left(a^4-a^2b^2+b^4\right)\left(a^8-a^4b^4+b^8\right)\)