How do you expand #(2y^3+x)^5#?

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Answer 1 · 2016-09-08T11:46:36

# x^5 +10y^3x^(4) + 40y^6x^(3)+ 80y^9x^(2)+80y^12x+32y^15#

Use the Binomial Theorem. First lets make a quick substitution

#(z+x)^5#, where #z=2y^3# now expand using the formula

#(y+x)^n =sum_(k=0)^n ("_k^n) y^kx^(n-k)#

# x^5 +5zx^(4) + 10z^2x^(3)+ 10z^3x^(2)+5z^4x+z^5#

now we can substitute back the z

# x^5 +5(2y^3)x^(4) + 10(2y^3)^2x^(3)+ 10(2y^3)^3x^(2)+5(2y^3)^4x+(2y^3)^5#

simplify

# x^5 +10y^3x^(4) + 40y^6x^(3)+ 80y^9x^(2)+80y^12x+32y^15#