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Exercise

(23a+4m)3\left(\frac{2}{3}a+4m\right)^3

Step-by-step Solution

1

The cube of a binomial (sum) is equal to the cube of the first term, plus three times the square of the first by the second, plus three times the first by the square of the second, plus the cube of the second term. In other words: (a+b)3=a3+3a2b+3ab2+b3=(23a)3+3(23a)2(4m)+3(23a)(4m)2+(4m)3=(a+b)^3=a^3+3a^2b+3ab^2+b^3 = (\frac{2}{3}a)^3+3(\frac{2}{3}a)^2(4m)+3(\frac{2}{3}a)(4m)^2+(4m)^3 =

(23a)3+12(23a)2m+2a(4m)2+(4m)3\left(\frac{2}{3}a\right)^3+12\left(\frac{2}{3}a\right)^2m+2a\left(4m\right)^2+\left(4m\right)^3
2

The power of a product is equal to the product of it's factors raised to the same power

827a3+12⋅(49)a2m+2⋅16am2+64m3\frac{8}{27}a^3+12\cdot \left(\frac{4}{9}\right)a^2m+2\cdot 16am^2+64m^3
3

Multiply 22 times 1616

827a3+12⋅(49)a2m+32am2+64m3\frac{8}{27}a^3+12\cdot \left(\frac{4}{9}\right)a^2m+32am^2+64m^3
4

Multiply the fraction and term in 12⋅(49)a2m12\cdot \left(\frac{4}{9}\right)a^2m

827a3+163a2m+32am2+64m3\frac{8}{27}a^3+\frac{16}{3}a^2m+32am^2+64m^3

Final answer to the exercise

827a3+163a2m+32am2+64m3\frac{8}{27}a^3+\frac{16}{3}a^2m+32am^2+64m^3

Try other ways to solve this exercise

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  • Product of Binomials with Common Term
  • FOIL Method
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  • Find the derivative
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  • Integrate by partial fractions
  • Integrate by substitution
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  • Integrate using tabular integration
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Main Topic: Special Products

Special products is the multiplication of algebraic expressions that follow certain rules and patterns, so you can predict the result without necessarily doing the multiplication.

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