Question

Show that -


${(a {x}^{2} + b {y}^{2} + c {z}^{2}) }^{ \frac{1}{3} } = {a}^{ \frac{1}{3} } + {b}^{ \frac{1}{3} } + {c}^{ \frac{1}{3} }$


If

$a {x}^{3} = b {y}^{3} = c {z}^{3}$

and

$\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 1$

​

Answer 1

★Given:

$a {x}^{3} = b {y}^{3} = c {z}^{3}$

and

$\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 1$

★To prove:

${(a {x}^{2} + b {y}^{2} + c {z}^{2}) }^{ \frac{1}{3} } = {a}^{ \frac{1}{3} } + {b}^{ \frac{1}{3} } + {c}^{ \frac{1}{3} }$

★Solution -

$Assume\\ \\ \\a {x}^{3} = b {y}^{3} = c {z}^{3} = K\\ \\ \\ a = \dfrac{K}{x^3} \\ \\ \\ b = \dfrac{K}{y^3} \\ \\ \\ c = \dfrac{K}{z^3}$

$by\: putting \:the \: value \:of \: a,b,c\\ \\ \\ {(a {x}^{2} + b {y}^{2} + c {z}^{2}) }^{ \frac{1}{3} } = {a}^{ \frac{1}{3} } + {b}^{ \frac{1}{3} } + {c}^{ \frac{1}{3} }$

$[(\dfrac{K}{\cancel{x^3}} \times \cancel{x^2}) + (\dfrac{K}{\cancel{y^3}} \times \cancel{y^2}) + (\dfrac{K}{\cancel{z^3}} \times \cancel{z^2})]^{\frac{1}{3}}$

$= (\dfrac{K}{x^3})^{\frac{1}{3}} + (\dfrac{K}{y^3})^{\frac{1}{3}} + (\dfrac{K}{z^3})^{\frac{1}{3}}$

$[\dfrac{K}{x} + \dfrac{K}{y} + \dfrac{K}{z}]^{\frac{1}{3}}$ $= \dfrac{K ^{\frac{1}{3}} }{x}+ \dfrac{K^{\frac{1}{3}} } {y} + \dfrac{K^{\frac{1}{3}} } {z}$

$[K (\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z})]^\frac{1}{3} = K^{\frac{1}{3}}(\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z})$

$\sf by\ putting\ the\ value\ of\ \frac{1}{x} +\frac{1}{y} +\frac{1}{z} \\ \\ \\ {[ K \times 1 ]}^{\frac{1}{3}} = [ {K}^{\frac{1}{3}} \times 1 ] \\ \\ {K}^{\frac{1}{3}} = {K}^{\frac{1}{3}} \\ \\ LHS = RHS$

${\pink{\boxed{Hence\: verified}}}$

Answer 2

GIVEN:

ax³ = by³ = cz³

1/x + 1/y + 1/z = 1

TO PROVE:

${(a {x}^{2} + b {y}^{2} + c {z}^{2}) }^{ \frac{1}{3} } = {a}^{ \frac{1}{3} } + {b}^{ \frac{1}{3} } + {c}^{ \frac{1}{3} }$

NEEDED TERMS:

1/x + 1/y + 1/z = 1

Take L.C.M

(yz + xz + xy) ÷ xyz = 1

yz + xz + xy = xyz

ax³ = by³

Take cube root on both sides

${a}^{ \frac{1}{3} } x = {b}^{ \frac{1}{3} } y \\ {b}^{ \frac{1}{3} } = \frac{{a}^{ \frac{1}{3} } x}{y}$

ax³ = cz³

Take cube root on both sides.

${a}^{ \frac{1}{3} } x = {c}^{ \frac{1}{3} } z \\ {c}^{ \frac{1}{3} } = \frac{{a}^{ \frac{1}{3} } x}{z}$

PROOF:

${(a {x}^{2} + b {y}^{2} + c {z}^{2}) }^{ \frac{1}{3} } = {a}^{ \frac{1}{3} } + {b}^{ \frac{1}{3} } + {c}^{ \frac{1}{3} } \\ Multiply \: by \: {(xyz)}^{ \frac{1}{3} } on \: both \: numerator \\ and \: denominator \: on \: left \: side \\ \frac{{(xyz)}^{ \frac{1}{3} }}{{(xyz)}^{ \frac{1}{3} }} {(a {x}^{2} + b {y}^{2} + c {z}^{2}) }^{ \frac{1}{3} } = {a}^{ \frac{1}{3} } + {b}^{ \frac{1}{3} } + {c}^{ \frac{1}{3} } \\ \\ Take \: L.H.S \\ = { \frac{(a {x}^{3}yz + b {y}^{3}xz + c {z}^{3} xy) }{ {(xyz)}^{ \frac{1}{3} } } }^{ \frac{1}{3} } \\ Substitute \: a {x}^{3} = b {y}^{3} = c {z}^{3} \\ = { \frac{(a {x}^{3}yz + a {x}^{3}xz + a {x}^{3} xy) }{ {(xyz)}^{ \frac{1}{3} } } }^{ \frac{1}{3} } \\ = {(a {x}^{3} )}^{ \frac{1}{3} } { (\frac{xy + yz + zx}{xyz}) }^{ \frac{1}{3} } \\ = {a}^{ \frac{1}{3} } x( {1)}^{ \frac{1}{3} } \\ = {a}^{ \frac{1}{3} } x \\ \\Take \: R.H.S \\ = {a}^{ \frac{1}{3} } + {b}^{ \frac{1}{3} } + {c}^{ \frac{1}{3} } \\ Substitute \: {b}^{ \frac{1}{3} } = \frac{ {a}^{ \frac{1}{3} }x }{y} and \: \: {c}^{ \frac{1}{3} } = \frac{ {a}^{ \frac{1}{3} }x }{z} \\ = {a}^{ \frac{1}{3}} + \frac{ {a}^{ \frac{1}{3} }x }{y} + \frac{ {a}^{ \frac{1}{3} }x }{z} \\ = {a}^{ \frac{1}{3} }( \frac{yz + zx + xy}{yz} ) \\ Equate \: L.H.S\: and \: R.H.S \\ {a}^{ \frac{1}{3} } x = {a}^{ \frac{1}{3} }( \frac{yz + zx + xy}{yz} ) \\ x = \frac{yz + zx + xy}{yz} \\ xyz = yz + xy + zx$

HENCE PROVED.