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

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

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