bạn tham khảo link này nhé:
https://olm.vn/hoi-dap/detail/101383958371.html
#hok tốt#
\(\left(a+b+c\right)^3+\left(a-b-c\right)^3-6a\left(b+c\right)^2\)
\(=a^3+b^3+c^3+a^3-b^3-c^3-6a\left(b^2+c^2\right)\)
\(=\left(a^3+a^3\right)+\left(b^3-b^3\right) +\left(c^3-c^3\right)-6a\left(b^2+c^2\right)\)
\(=2a^3-6a\left(b^2+c^2\right)\)
\(=2a^2\cdot a-6a\left(b^2+c^2\right)\)
\(=a\left[2a^2-6\left(b^2+c^2\right)\right]\)
\(\text{Chắc là vậy !}\)
Link nè :
https://olm.vn/hoi-dap/detail/101383958371.html
k mk nhé mk hông giải nên mk cho bn link rùi nè
\(\left(a+b+c\right)^3+\left(a-b-c\right)^3-6a\left(b+c\right)^2\)
\(=\left[a+\left(b+c\right)\right]^3+\left[a-\left(b+c\right)\right]^3-6a\left(b+c\right)^2\)
\(=a^3+3a^2\left(b+c\right)+3a\left(b+c\right)^2+\left(b+c\right)^3+a^3-3a^2\left(b+c\right)\)
\(+3a\left(b+c\right)^2-\left(b+c\right)^3-6a\left(b+c^2\right)\)
\(=\left(a^3+a^3\right)+\left[3a^2\left(b+c\right)-3a^2\left(b+c\right)\right]+\left[3a\left(b+c\right)^2+3a\left(b+c\right)^2\right]\)
\(+\left[\left(b+c\right)^3-\left(b+c\right)^3\right]-6a\left(b+c\right)^2\)
\(=2a^3+0+6a\left(b+c\right)^2+0-6a\left(b+c\right)^2\)
\(=2a^3+6a\left(b+c\right)^2-6a\left(b+c\right)^2\)
\(=2a^3+0=2a^3\)
