Question

Math

Posted 5 months ago

`The number $(\sqrt{2}+\sqrt{3})^{3}$ can be written in the form $a \sqrt{2}+b \sqrt{3}+c \sqrt{6}$, where $a, b$, and $c$ are integers. What is $a+b+c$ ?`

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Answer from Sia

Posted 5 months ago

Solution by Steps

step 1

To expand $(\sqrt{2}+\sqrt{3})^3$, we will use the binomial expansion formula $(a+b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$

step 2

Let $a=\sqrt{2}$ and $b=\sqrt{3}$. Compute $a^3$, $3a^2b$, $3ab^2$, and $b^3$

step 3

Calculating $a^3 = (\sqrt{2})^3 = 2\sqrt{2}$

step 4

Calculating $3a^2b = 3(\sqrt{2})^2\sqrt{3} = 3 \cdot 2\sqrt{3} = 6\sqrt{3}$

step 5

Calculating $3ab^2 = 3\sqrt{2}(\sqrt{3})^2 = 3\sqrt{2} \cdot 3 = 9\sqrt{2}$

step 6

Calculating $b^3 = (\sqrt{3})^3 = 3\sqrt{3}$

step 7

Adding the results from steps 3 to 6, we get $(\sqrt{2}+\sqrt{3})^3 = 2\sqrt{2} + 6\sqrt{3} + 9\sqrt{2} + 3\sqrt{3}$

step 8

Combine like terms to get $a\sqrt{2} + b\sqrt{3} = (2+9)\sqrt{2} + (6+3)\sqrt{3} = 11\sqrt{2} + 9\sqrt{3}$

step 9

There is no $\sqrt{6}$ term, so $c=0$

step 10

The values of $a$, $b$, and $c$ are $11$, $9$, and $0$ respectively

step 11

To find $a+b+c$, we add $11+9+0$

step 12

The sum is $11+9+0 = 20$

Answer

$a+b+c = 20$

Key Concept

Binomial Expansion

Explanation

The binomial expansion formula is used to expand expressions of the form $(a+b)^n$. In this case, we expanded $(\sqrt{2}+\sqrt{3})^3$ and combined like terms to find the coefficients $a$, $b$, and $c$.

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