Super confused over simple unit conversion from centimeters to cubic centimeters

I'm stuck in a really stupid question. It seems like the meaning of "centi" is chancing meaning. Because if:

$$r = 5 cm = 5 \cdot 10^{-2} m$$

so $$c = 10^{-2}$$

But say i need to take the cubic root of r:

$$r^3 = 125 cm^3$$, this is true. But in my head it should be:

$$r^3 = (5cm)^3 = 5^3 c^3m^3 = 125\cdot10^{-6} m^3$$

So while the true answer is centi meters^3, the calculation only makes sencse if the answer is in micro meters^3...

Would love your inupt i am going nuts

• I don't understand your last line. Everything looks fine to me? 125 cm^3 = 125*10^(-6) m^3 as stated correctly. – Noldig Feb 21 at 18:27
• Do you understand that if there are $3$ feet in a yard, there are $9$ square feet in a square yard? – knzhou Feb 21 at 18:43
• You are getting micro (cubic meters). you are not getting (micrometers) cubed (which would be 10 to the -18). – ohwilleke Feb 21 at 18:45
• @Noldig last line is what i think should be right. knzhou thank you, but i understand now that SI prefixing are combining, like what indiaria wrote – Pernk Dernets Feb 21 at 18:51
• a centi (cubic meter) is not the same as a cubic (centi meter). – ja72 Feb 21 at 21:42

SI prefixes are combining, so when you say $$\text{cm}^3$$, you're actually saying $$(\text{cm})^3 = (10^{-2} \text{m})^3 = 10^{-6} \text{m}^3$$, similar to how $$\mathrm{d}^2y/\mathrm{d}x^2$$ refers to $$\mathrm{d}^2y/(\mathrm{d}x)^2$$ and not $$\mathrm{d}^2y/\mathrm{d}(x^2)$$. It might help if you if you think it as a different unit "$$\text{cm}$$" rather than "$$\text{c}\cdot\text{m}$$". If in doubt you can always expand and factorize it:
$$140 \text{km}^2 = 140\times(10^3 \text{m})^2 = 1.4\times 10^8 \text{m} \\ = 1.4\times 10^{8}\times (10^{6})^2\times(10^{-6}\text{m})^2 \\ = 1.4\times 10^{8+6\times 2}(10^{-6}\text{m})^2 = 1.4\times 10^{20} \text{µm}^2$$
A cubic micron is $$(10^{-6})^3 = 10^{-18}$$ cubic meters. Your mistake was saying a micro-cubic-meter was a cubic micrometer (cubic micron).