1 Answer

Ishikawa Yoshi · Answer 1 · 2023-10-30T13:45:33+0000

The binomial can be expanded using the formula:

$(a+b)^n=\sum ^n_(k\mathop=0)(n!)/((n-k)!k!)a^(n-k)b^k$

For the binomial (x³+2/x)⁸, we have:

n = 8

a = x³

b = 2/x

So, the constant term is the one for which:

$\mleft(x^3\mright)^(8-k)\mleft((2)/(x)\mright)^k=\text{ constant}$

This happens for k = 6:

$a^(n-k)b^k=(x^3)^(8-6)\mleft((2)/(x)\mright)^6=x^6\cdot(2^6)/(x^6)=2^6=64$

The, for k = 6, we have:

$(n!)/((n-k)!k!)=(8!)/(2!6!)=(8\cdot7\cdot6!)/(2\cdot1\cdot6!)=(56)/(2)=28$

Thus, the constant term in the expansion of the binomial is:

$(n!)/((n-k)!k!)a^(n-k)b^k=28\cdot64=1792$

Therefore, the answer is 1792.

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