Cube Roots

l learnt his method. 3~5 + 8~4 + 1~2 = 131.

i would read the note under the problem before attempting; it was very helpful! first, break the numbers down. so let 42875 be 42 and 875. so, since the 875 ends in a 5, we can say that the root of the entire number ends in a 5 (5 5 5=125). next, we can break the 42 down to the cube root 3. basically, we just solve for 30^3 and then put the 5 in because it ends in a 5. we do the same thing for the second number by separating 592704 into 592 and 704 (so 8^3 and 4 to make 84). the final number is smaller so we can just let the first digit be 10 because 10 10 10=1000 and let the second digit be 2 because 2^3=8. finally, we add up what we have to get our answer: 35+84+131. the same ideas are used for mental math and number crunching. my first try at explaining answers so i hope that was clear enough!

42875 = 35^3

592704 = 84^3

1728 = 12^3

Then

cubic root of 42875 +cubic root of 592704 + cubic root of 1728 =

35 + 84 + 12 = 131

cube of certain numbers up to 15 you must know.

Similarly you must know cube of numbers ending with 0, 1, 4, 5 , 6 always ends with 0, 1, 4, 5, 6 digits respectively

similarly cube of numbers ending with 2 and 8 end 8 with 2 respectively

similarly cube of numbers ending with 3 and 7 end 7 with 3 respectively

still to find cube root of perfect cube number, make groups from the end

say 1728 groups are 1 and 728 so your cube root ends with 2

consider first group consist 1 (if not 1) find a cube number ending with 1 or the required number. in our case it is 1

so approximate cube root is 12 confirm by working (you must know shortcut methods to find square of numbers (learn vedic maths.

in case of 42 875 cube root ends with 5 and cube root 42 is 3 (3^3 = 27 and 4^4 = 64) 42^1/3 lies between 3 and 4

so approximate value is 35 confirm

In case of 592 704 cube root ends with 4 and 8^3 < 592 < 9^3

so approximate answer is 84 confirm it

add all values = 131

Using prime factorization it's easy $\sqrt[3]{42875}+\sqrt[3]{592704}+\sqrt[3]{1728} = \sqrt[3]{5^3*7^3}+\sqrt[3]{2^6*3^3*7^3}+\sqrt[3]{2^6*3^3}$

$=5*7+2^2*3*7+2^2*3=35+84+12 =\fbox{131}$

35+84+12 = 131

30^3<42875<40^3 --> 42875 = 35^3 80^3<592704<90^3 --> 592704 = 84^3 10^3<1728<20^3 --> 1728 = 12^3

answer = 35+84+12=131

35+84+12=131