Question

64 x cube - 125 z cube if 4x-5z=16 and xz=12?
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Answer 1

$ANSWER</p><p></p><p>Given, 4x - 5z = 16 and xz = 12</p><p>Since, (a−b)2=a2−2ab+b2</p><p>⇒(4x−5z)2=16x2−40xz+25z2</p><p>put 4x - 5z = 16 and xz = 12</p><p>⇒(16)2=16x2−40(12)+25z2</p><p>⇒16x2+25z2=256+480</p><p>⇒16x2+25z2=736          .....(1)</p><p> Given, 64x3−125z3=(4x)3−(5z)3</p><p>Since,a3−b3=(a−b)(a2+b2+ab)</p><p>⇒(4x)3−(5z)3=(4x−5z)(16x2+25z2+20xz)</p><p>By putting 4x - 5z = 16 , xz = 12 and using (1). we get,</p><p>⇒(4x)3−(5z)3=(16)(736+20×12)</p><p>⇒(4x)3−(5z)3=(16)(976)</p><p>⇒(4x)3−(5z)3=15616</p><p>Option D is correct.</p><p></p><p>$

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