Câu hỏi của Ngô Huy Hoàng

Question

Cho $a^3-3ab^2=5$ và $b^3-3a^2b=10$. Tính $M=a^2+b^2$

Gia Hân · Accepted Answer

Ta có $\left(a^3-3ab^2\right)^2$ =$a^6-6a^4b^2+9a^2b^4=25$

$\left(b^3-3a^2b\right)^2=b^6-6a^2b^4+9a^4b^2=100$

$=>\left(a^3-3a^2b\right)^2-\left(b^3-3a^2b\right)^2=a^6-6a^4b^2+9a^2b^4+b^6-6a^2b^4+9a^4b^2=125$

$< =>a^6+3a^4b^2=3a^2b^4+b^6=125$

$< =>\left(a^2+b^2\right)^3=125$

$=>a^2+b^2=5$

Câu trả lời:

Ta có $\left(a^3-3ab^2\right)^2$ =$a^6-6a^4b^2+9a^2b^4=25$

$\left(b^3-3a^2b\right)^2=b^6-6a^2b^4+9a^4b^2=100$

$=>\left(a^3-3a^2b\right)^2-\left(b^3-3a^2b\right)^2=a^6-6a^4b^2+9a^2b^4+b^6-6a^2b^4+9a^4b^2=125$

$< =>a^6+3a^4b^2=3a^2b^4+b^6=125$

$< =>\left(a^2+b^2\right)^3=125$

$=>a^2+b^2=5$