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<!DOCTYPE html>
<html data-require="math graphie">
<head>
<title>Cube roots</title>
<script src="../khan-exercise.js"></script>
</head>
<body>
<div class="exercise">
<div class="vars">
<var id="cx">0</var>
<var id="y">0</var>
<var id="N">randRange( 2, 12 )</var>
<var id="F_N">getPrimeFactorization( N )</var>
</div>
<div class="hints">
<p>
If you can't think of that number, you can break down <code><var>Q</var></code> into
its prime factorization and look for equal groups of numbers.
</p>
<div>
<p>Let's draw a factor tree.</p>
<div class="graphie" id="factor-tree">
init({
range: [ [-1, FACTORIZATION.length + 2], [ -2 * FACTORIZATION.length - 1, 1] ],
scale: [30, 30]
});
label( [cx + 1, y], curr );
</div>
</div>
<div class="graphie" data-update="factor-tree" data-each="FACTORIZATION as factor">
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 );
</div>
<div class="graphie" data-update="factor-tree">
circle( [cx + 1, y], 0.5);
</div>
<p>So the prime factorization of <code><var>Q</var></code> is <code><var>PRIMES.join( "\\times " )</var></code>.</p>
</div>
<div class="problems">
<div id="cube">
<div class="vars">
<var id="Q">N * N * N</var>
<var id="PRIMES">getPrimeFactorization( Q )</var>
<var id="FACTORIZATION">PRIMES.slice( 0, PRIMES.length - 1 )</var>
<var id="curr">Q</var>
</div>
<p class="question">What is <code>\sqrt[3]{<var>Q</var>}</code>?</p>
<p class="solution"><var>N</var></p>
<div class="hints" data-apply="prependContents">
<p>
<code>\sqrt[3]{<var>Q</var>}</code> is the number that, when
multiplied by itself three times, equals <code><var>Q</var></code>.
</p>
</div>
<div class="hints" data-apply="appendContents">
<p>We're looking for <code>\sqrt[3]{<var>Q</var>}</code>, so we want to split the prime factors into three identical groups.</p>
<div data-if="PRIMES.length === 3" data-unwrap>
<p>We only have three prime factors, and we want to split them into three groups, so this is easy.</p>
<p><code><var>Q</var> = <var>PRIMES.join( "\\times " )</var></code>, so <code><var>N</var>^3 = <var>Q</var></code>.</p>
</div><div data-else data-unwrap>
<div>
<p>Notice that we can rearrange the factors like so:</p>
<p><code><var>Q</var> = <var>PRIMES.join( "\\times " )</var> = \left(<var>[ F_N.join( "\\times " ), F_N.join( "\\times " ), F_N.join( "\\times ") ].join( "\\right)\\times\\left(" )</var>\right)</code><p>
</div>
<p data-if="F_N.length > 1">
So <code>\left(<var>F_N.join( "\\times " )</var>\right)^3 = <var>N</var>^3 = <var>Q</var></code>.
</p><p data-else>
So <code><var>N</var>^3 = <var>Q</var></code>.
</p>
</div>
<p class="final_answer">So <code>\sqrt[3]{<var>Q</var>}</code> is <code><var>N</var></code>.</p>
</div>
</div>
</div>
</div>
</body>
</html>
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