# Khan/khan-exercises

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 Cube roots
0 0 randRange( 2, 12 ) getPrimeFactorization( N )

If you can't think of that number, you can break down Q into its prime factorization and look for equal groups of numbers.

Let's draw a factor tree.

init({ range: [ [-1, FACTORIZATION.length + 2], [ -2 * FACTORIZATION.length - 1, 1] ], scale: [30, 30] }); label( [cx + 1, y], curr );
path( [ [cx + 1, y - 0.5], [cx, y - 1.5] ] ); path( [ [cx + 1, y - 0.5], [cx + 2, y - 1.5] ] ); y -= 2; cx += 1; curr = curr / factor; label( [cx - 1, y], factor ); circle( [cx - 1, y], 0.5); label( [cx + 1, y], curr );
circle( [cx + 1, y], 0.5);

So the prime factorization of Q is PRIMES.join( "\\times " ).

N * N * N getPrimeFactorization( Q ) PRIMES.slice( 0, PRIMES.length - 1 ) Q

\Large{\sqrt[3]{Q} = \text{?}}

N

\sqrt[3]{Q} is the number that, when multiplied by itself three times, equals Q.

We're looking for \sqrt[3]{Q}, so we want to split the prime factors into three identical groups.

We only have three prime factors, and we want to split them into three groups, so this is easy.

Q = PRIMES.join( "\\times " ), so N^3 = Q.

Notice that we can rearrange the factors like so:

Q = PRIMES.join( "\\times " ) = \left([ F_N.join( "\\times " ), F_N.join( "\\times " ), F_N.join( "\\times ") ].join( "\\right)\\times\\left(" )\right)

So \left(F_N.join( "\\times " )\right)^3 = N^3 = Q.

So N^3 = Q.

So \sqrt[3]{Q} is N.