Question

CH: Ratio
1) If x/a²-b² = y/b²-c²= z/c²-a², then prove that x + y + z = 0.
2) If x/a=y/b=z/c, the prove that each ratio is equal to (3x³-11y³+13z³/3a³-11b³+13c³)^1/3.
3) If x:y:z = 3:4:5 and x³+y³+z³ = 1728, then find (x-y+z).

Give correct answer please. I will mark as brainliest. Please

Answer 1

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$\sf{1)}$ $\large\sf{\frac{x}{ {a}^{2} - {b}^{2} } = \frac{y}{ {b}^{2} - {c}^{2} } = \frac{z}{ {c}^{2} - {a}^{2} },}$ $\sf{\:then\:prove\:that\:x+y+z=0.}$

${\bf{\red{\underline{\underline{Solution:}}}}}$

Let's assume that, $\large\sf{\frac{x}{ {a}^{2} - {b}^{2} } = \frac{y}{ {b}^{2} - {c}^{2} } = \frac{z}{ {c}^{2} - {a}^{2} }=k}$

$\implies{\sf{x=k(a^2-b^2),\:y=k(b^2-c^2),\:z=k(c^2-a^2)}}$

$\implies{\sf{x+y+z=k(\cancel{a^2}\cancel{-b^2}+\cancel{b^2}\cancel{-c^2}+\cancel{c^2}\cancel{-a^2})}}$

$\implies{\sf{x+y+z=k\times 0}}$

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎$\underline{\boxed{\sf{\therefore x+y+z=0}}}$

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$\sf{2)\:If}$ $\large\sf{ \frac{x}{a} = \frac{y}{b} = \frac{z}{c} ,}$ $\sf{then\:prove\:that\:each\:ratio\:is\:equal\:to}$ $\large\sf{{\bigg( \frac{ {3x}^{3} - {11y}^{3} + {13z}^{3} }{ {3a}^{3} - {11b}^{3} + {13c}^{3} } \bigg)}}^{\frac{1}{3}}$.

${\bf{\red{\underline{\underline{Solution:}}}}}$

Let, $\large\sf{\frac{x}{a} = \frac{y}{b} = \frac{z}{c}=k}$

$\implies{\sf{x=ak,\:y=bk\:and\:z=ck}}$

$\large\sf{\therefore {\bigg( \frac{ {3x}^{3} - {11y}^{3} + {13z}^{3} }{ {3a}^{3} - {11b}^{3} + {13c}^{3} } \bigg)}}^{\frac{1}{3}}={\bigg(\frac{ {3(ak)}^{3} - {11(bk)}^{3} + {13(ck)}^{3} }{ {3a}^{3} - {11b}^{3} + {13c}^{3} } \bigg)}^{\frac{1}{3}}={\bigg(\frac{ {k}^{3} ( {3a}^{3} - {11b}^{3} + {13c}^{3}) }{( {3a}^{3} - {11b}^{3} + {13c}^{3}) }\bigg)} ^{\frac{1}{3}}=k$

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎$\underline{\boxed{\sf{Hence,\:proved!}}}$

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$\sf{3)\:If\:x:y:z=3:4:5\:and\:x^3+y^3+z^3=1728,\:then\:find\:(x-y+z).}$

${\bf{\red{\underline{\underline{Solution:}}}}}$

Let, $\sf{x=3a}$, $\sf{y=4a}$ and $\sf{z=5a}$.

Given, $\sf{x^3+y^3+z^3=1728}$

$\implies{\sf{(3a)^3+(4a)^3+(5a)^3=1728}}$

$\implies{\sf{216a^3=1728}}$

$\implies{\sf{a^3=8}}$

$\implies{\sf{a=2}}$

$\sf{\therefore x-y+z= 3a-4a+5a}$

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎$\sf{=4a}$

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎ ‎$\sf{=4\times 2}$

‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎$\sf{=\red{\underline{8}}}$

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Answer 2

Answer:

$) \large\sf{\frac{x}{ {a}^{2} - {b}^{2} }</p><p> = \frac{y}{ {b}^{2} - {c}^{2} } </p><p>= \frac{z}{ {c}^{2} - {a}^{2} },}a2−b2x=b2−c2y=c2−a2z,</p><p> \sf{\:then\:prove\:that\:x+y+z=0.}thenprovethatx+y+z=0.</p><p></p><p></p><p>{\bf{\red{\underline{\underline{Solution:}}}}}Solution:</p><p></p><p>Let's assume that, \large\sf{\frac{x}{ {a}^{2} - {b}^{2} }</p><p> = \frac{y}{ {b}^{2} - {c}^{2} }</p><p> = \frac{z}{ {c}^{2} - {a}^{2} </p><p>=k}a2−b2x</p><p>=b2−c2y</p><p>=c2−a2z</p><p>=k</p><p></p><p>\implies{\sf{x=k(a^2-b^2),\:y=</p><p>k(b^2-c^2),\:z=k(c^2-a^2)}}⟹x=k(a2−b2),y</p><p>=k(b2−c2),z</p><p>=k(c2−a2)</p><p></p><p>\implies{\sf{x+y+z=k(\cancel{a^2}\cancel{-b^2}+\cancel{b^2}\cancel{-c^2}+\cancel{c^2}\cancel{-a^2})}}⟹x+y+z=k(a2−b2+b2−c2+c2−a2)</p><p></p><p>\implies{\sf{x+y+z=k\times 0}}⟹x+y+z=k×0</p><p></p><p>‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎\underline{\boxed{\sf{\therefore x+y+z=0}}}∴x+y+z=0</p><p></p><p></p><p>$

$)If \large\sf{ \frac{x}{a}</p><p> = \frac{y}{b} = \frac{z}{c} ,}ax=by=cz, </p><p>\sf{then\:prove\:that\:each\:ratio\:is\:equal\:to}</p><p></p><p>thenprovethateachratioisequalto \large\sf{{\bigg( \frac{ {3x}^{3} - {11y}^{3} + {13z}^{3} }{ {3a}^{3} - {11b}^{3} + {13c}^{3} } \bigg)}}^{\frac{1}{3}}(3a3−11b3+13c33x3−11y3+13z3)31 .</p><p></p><p>{\bf{\red{\underline{\underline{Solution:}}}}}Solution:</p><p></p><p>Let, \large\sf{\frac{x}{a} = \frac{y}{b} = \frac{z}{c}=k}ax=by=cz=k</p><p></p><p>\implies{\sf{x=ak,\:y=bk\:and\:z=ck}}⟹x=ak,y=bkandz=ck</p><p></p><p>\large\sf{\therefore {\bigg( \frac{ {3x}^{3} - {11y}^{3} + {13z}^{3} }{ {3a}^{3} - {11b}^{3} + {13c}^{3} } \bigg)}}^{\frac{1}{3}}={\bigg(\frac{ {3(ak)}^{3} - {11(bk)}^{3} + {13(ck)}^{3} }{ {3a}^{3} - {11b}^{3} + {13c}^{3} } \bigg)}^{\frac{1}{3}}={\bigg(\frac{ {k}^{3} ( {3a}^{3} - {11b}^{3} + {13c}^{3}) }{( {3a}^{3} - {11b}^{3} + {13c}^{3}) }\bigg)} ^{\frac{1}{3}}=k∴(3a3−11b3+13c33x3−11y3+13z3)31=(3a3−11b3+13c33(ak)3−11(bk)3+13(ck)3)31=((3a3−11b3+13c3)k3(3a3−11b3+13c3))31=k</p><p></p><p>‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎\underline{\boxed{\sf{Hence,\:proved!}}}Hence,proved!</p><p></p><p>$

$)Ifx:y:z=3:4:5andx3+y3+z3=1728,thenfind(x−y+z).</p><p></p><p>{\bf{\red{\underline{\underline{Solution:}}}}}Solution:</p><p></p><p>Let, \sf{x=3a}x=3a , \sf{y=4a}y=4a and \sf{z=5a}z=5a .</p><p></p><p>Given, \sf{x^3+y^3+z^3=1728}x3+y3+z3=1728</p><p></p><p>\implies{\sf{(3a)^3+(4a)^3+(5a)^3=1728}}⟹(3a)3+(4a)3+(5a)3=1728</p><p></p><p>\implies{\sf{216a^3=1728}}⟹216a3=1728</p><p></p><p>\implies{\sf{a^3=8}}⟹a3=8</p><p></p><p>\implies{\sf{a=2}}⟹a=2</p><p></p><p>\sf{\therefore x-y+z= 3a-4a+5a}∴x−y+z=3a−4a+5a</p><p></p><p>‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎\sf{=4a}=4a</p><p></p><p>‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎ ‎\sf{=4\times 2}=4×2</p><p></p><p>‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎ ‎‎\sf{=\red{\underline{8}}}=8</p><p></p><p>$