I don't understand "collecting radical terms"

znick46

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Dec 19, 2013
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Example:
Step 0: 10y^3√2y +7^3√2y
Step 1: (10y+7) ^3√2y

By memory I know I can add 10y + 7 and combine the two "^3√2y" to be one(as illustrated in step 1).

I don't understand why, please explain intuitively how this works.

Thank You
 
Example:
Step 0: 10y^3√2y +7^3√2y
Step 1: (10y+7) ^3√2y

By memory I know I can add 10y + 7 and combine the two "^3√2y" to be one(as illustrated in step 1).

I don't understand why, please explain intuitively how this works.

Thank You

Is this exercise supposed to be \(\displaystyle \ \ 10y\sqrt[3]{2y} \ + \ 7\sqrt[3]{2y} \ \ ?\)
 
Actually, in general, a^x + b^x does not equal (a+b)^x. Try it, for example, with a = 3, b = 4, and x = 2...doesn't work.

This, however, is true: (a^x)(b^x) = (ab)^x.
 
Actually, in general, a^x + b^x does not equal (a+b)^x. Try it, for example, with a = 3, b = 4, and x = 2...doesn't work.

This, however, is true: (a^x)(b^x) = (ab)^x.

MathMathter, if you read what I asked of the original poster, I don't think he's making that claim.

He appears to be attempting communicate "^3√2y" to mean the cube root of 2y.

I was waiting for a response from him about that.
 
Example:
Step 0: 10y^3√2y +7^3√2y
Step 1: (10y+7) ^3√2y

By memory I know I can add 10y + 7 and combine the two "^3√2y" to be one(as illustrated in step 1).

I don't understand why, please explain intuitively how this works.

Thank You
Please read the preceding posts carefully.

I am going to assume that you meant \(\displaystyle 10y\sqrt[3]{2y} + 7\sqrt[3]{2y}.\)

You should type that in line as 10y(2y)^(1/3) + 7(2y)^(1/3).

\(\displaystyle \sqrt[3]{2y}\) is just a number. For example if y = 4 then \(\displaystyle \sqrt[3]{2y} = \sqrt[3]{2 * 4} = \sqrt[3]{8} = 2.\)

Of course, for most values of y, the cube root will be a long messy decimal, but it still is just a number.

Think about 10yz + 7z = z(10y + 7) = (10y + 7)z. That is true for any number so it is true for cube roots, which are numbers.

Example

\(\displaystyle y = 108 \implies 10y\sqrt[3]{2y} + 7\sqrt[3]{2y} = 10 * 108 * \sqrt[3]{216} + 7\sqrt[3]{216} = 1080 * 6 + 7 * 6 = 6480 + 42 = 6522.\)

\(\displaystyle y = 108 \implies 10y\sqrt[3]{2y} + 7\sqrt[3]{2y} = (10y + 7)\sqrt[3]{2y} = (10 * 108 + 7)\sqrt[3]{2 *108} = (1080 + 7)\sqrt[3]{216} = 1087 * 6 = 6522.\)

Does that help?
 
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