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ICSE Class X Math 2016 Paper

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Question : 33 of 57
Marks: +1, -0
If   xa=  yb=  zc\; \frac{x}{a} = \; \frac{y}{b} = \; \frac{z}{c} show that   x3a3+  y3b3+  z3c3=  3xyzabc\; \frac{x^3}{a^3} + \; \frac{y^3}{b^3} + \; \frac{z^3}{c^3} = \; \frac{3xyz}{abc}.
Solution:  
Given:   xa=  yb=  zc=k\; \frac{x}{a} = \; \frac{y}{b} = \; \frac{z}{c} = k (say)
  ⇒    x=ak,y=bk,z=ck\; \Rightarrow \; \; x = a k, y = b k, z = c k
     L.H.S.   =  x3a3+  y3b3+  z3c3\; \; \text{ L.H.S. } \; = \; \frac{x^3}{a^3} + \; \frac{y^3}{b^3} + \; \frac{z^3}{c^3}
    =  a3k3a3+  b3k3b3+  c3k3c3\; \; = \; \frac{a^3 k^3}{a^3} + \; \frac{b^3 k^3}{b^3} + \; \frac{c^3 k^3}{c^3}
    =k3+k3+k3=3k3\; \; = k^3 + k^3 + k^3 = 3 k^3
    =3k⋅k⋅k\; \; = 3 k \cdot k \cdot k
    =3  xa  yb  zc\; \; = 3 \; \frac{x}{a} \; \frac{y}{b} \; \frac{z}{c}
(∵x=ak,y=bk,z=ck)( \because x = a k, y = b k, z = c k )
    =  3xyzabc=   R.H.S.   \; \; = \; \frac{3xyz}{abc} = \; \text{ R.H.S. } \;
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