How do you use the Binomial Theorem to expand #(3x^2 - (5/x))^3 #?

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Answer 1 · 2016-09-05T06:16:00

#(3x^2-5/x)^3=27x^6-135x^3+225-125/x^3#.

#(a+b)^n=""_nC_0*a^(n-0)*b^0+""_nC_1*a^(n-1)*b^1+""_nC_2*a^(n-2)*b^2+""_nC_3*a^(n-3)*b^3+...+""_nC_n*a^(n-n)*b^n#.

#:.(a-b)^3=""_3C_0*a^(3-0)*(-b)^0+""_3C_1*a^(3-1)*(-b)^1+""_3C_2*a^(3-2)*(-b)^2+""_3C_3*a^(3-3)*(-b)^3#.

Knowing that, #""_3C_0=""_3C_3=1, and, ""_3C_1=""_3C_2=3#,

#(a-b)^3=a^3-3a^2b+3ab^2-b^3.....................(star)#

Letting, #a=3x^2, and, b=5/x" in "(star)#, we get,

#(3x^2-5/x)^3=27x^6-3(9x^4)(5/x)+3(3x^2)(5/x)^2-125/x^3#

#=27x^6-135x^3+225-125/x^3#.

Enjoy maths.!