Question

$\sf{If}{ \tt{ \: \dfrac{ {x}^{2} - yz }{a} = \dfrac{ {y}^{2} - zx}{b} = \dfrac{ {z}^{2} - xy }{c} \: \cancel{ = } \: 0}}$
$\\ \sf{show \: that - } \tt{(a + b + c)(x + y + z) = ax + by + cz}$
$\\ \red{ \sf{ \rightarrow{no \: spam}}}$
​

Answer 1

Topic :-

$\mathtt{Algebra}$

Given :-

$\mathtt{\dfrac{x^2-yz}{a}=\dfrac{y^2-zx}{b}=\dfrac{z^2-xy}{c}\neq 0}$

To Prove :-

$\mathtt{(a+b+c)(x+y+z)=ax+by+cz}$

Solution :-

$\mathtt{Let\:us\:assume\:that\:given\:ratios\:equals\:to\:'k'.}$

$\mathtt{k\neq 0}$

$\mathtt{\dfrac{x^2-yz}{a}=\dfrac{y^2-zx}{b}=\dfrac{z^2-xy}{c}=k}$

We may write,

$\mathtt{\dfrac{x^2-yz}{a}=k}$

$\mathtt{\dfrac{x^2-yz}{k}=a}$

$\mathtt{\dfrac{y^2-zx}{b}=k}$

$\mathtt{\dfrac{y^2-zx}{k}=b}$

$\mathtt{\dfrac{z^2-xy}{c}=k}$

$\mathtt{\dfrac{z^2-xy}{k}=c}$

Solving LHS,

$\mathtt{(a+b+c)(x+y+z)}$

Substitute values of a, b and c,

$\mathtt{\left(\dfrac{x^2-yz}{k}+\dfrac{y^2-zx}{k}+\dfrac{z^2-xy}{k}\right)(x+y+z)}$

$\mathtt{\left( \dfrac{x^2-yz+y^2-zx+z^2-xy}{k}\right )(x+y+z)}$

$\mathtt{\dfrac{(x^2+y^2+z^2-xy-yz-zx)(x+y+z)}{k}}$

$\mathtt{\dfrac{x^3+y^3+z^3-3xyz}{k}}$

$\mathtt{(\because (p+q+r)(p^2+q^2+r^2-pq-qr-rp)=p^3+q^3+r^3-3pqr)}$

Solving RHS,

$\mathtt{ax+by+cz}$

Substitute values of a, b and c,

$\mathtt{\left(\dfrac{x^2-yz}{k}\right)x+\left(\dfrac{y^2-zx}{k}\right)y+\left(\dfrac{z^2-xy}{k}\right)z}$

$\mathtt{\dfrac{x^3-xyz}{k}+\dfrac{y^3-xyz}{k}+\dfrac{z^3-xyz}{k}}$

$\mathtt{\dfrac{x^3-xyz+y^3-xyz+z^3-xyz}{k}}$

$\mathtt{\dfrac{x^3+y^3+z^3-3xyz}{k}}$

We observe that LHS = RHS !!

Hence, Proved !!

Answer 2

$\large{\dag{\underline{\underline{\sf{\pmb{Question: - }}}}}}$

$\sf{If \: \dfrac{ {x}^{2} - yz }{a} = \dfrac{ {y}^{2} - zx}{b} = \dfrac{ {z}^{2} - xy }{c} \: \cancel{ = } \: 0}$

$\sf{show \: that : - (a + b + c)(x + y + z) = ax + by + cz}$

$\large{\dag{\underline{\underline{\sf{\pmb{Required \: Answer: - }}}}}}$

$\\ \dag{\underline{\underline{\sf{\pmb{Given: - }}}}}$

$\\ \sf{\rightarrow{ \: \dfrac{ {x}^{2} - yz }{a} = \dfrac{ {y}^{2} - zx}{b} = \dfrac{ {z}^{2} - xy }{c} \: \cancel{ = } \: 0}}$

$\\ \dag{\underline{\underline{\sf{\pmb{To \: Do: - }}}}}$

$\\ \sf{\rightarrow{Show \: that : - (a + b + c)(x + y + z) = ax + by + cz}}$

$\\\dag{\underline{\underline{\sf{\pmb{Solution: - }}}}}$

Let us assume that :-

$\sf{\dfrac{ {x}^{2} - yz}{a} = \dfrac{ {y}^{2} - zx }{b} = \dfrac{ {z}^{2} - xy }{c} = \bold{k}}$

Therefore,

$\sf{\therefore{\dfrac{ {x}^{2} - yz}{a} = k}}$

$\\ \sf{\implies{\blue{ {x}^{2} - yz = ak - - - - - - - - - - - (1)}}}$

Similarly:-

$\sf{\blue{ {y}^{2} - zx = bk - - - - - - - - - - - (2)}}$

$\sf{\blue{ {z}^{2} - xy = ck- - - - - - - - - - - (3)}}$

Now:-

$\sf{\bold{(1) \times x + (2) \times y + (3) \times z \implies}}$

$\\ \sf{\bold{( {x}^{2} - yz)x + ( {y}^{2} - zx)y + ( {z}^{2} - xy)z = ak \times x + bk \times y + ck \times z}}$

$\\ \sf{\implies{{x}^{3} - xyz + {y}^{3} - xyz + {z}^{3} - xyz = k(ax + by + cz)}}$

$\\ \sf{\implies{{x}^{3} + {y}^{3} + {z}^{3} - 3xyz = k(ax + by + cz)}}$

Now, we know that:-

$\\ \underline{\boxed{\pmb{ {a}^{3} + {b}^{3} + {c}^{3} - 3abc = (a + b + c)( {a}^{2} + {b}^{2} + {c}^{2} - ab - bc - ca)}}}$

So, applying the formula we get:-

$\\ \sf{\implies{(x + y + z)( {x}^{2} + {y}^{2} + {z}^{2} - xy - yz - zx) = k(ax + by + cz)}}$

$\\ \sf{\implies{(x + y + z)( {x}^{2} - yz + {y}^{2} - zx + {z}^{2} - xy) = k(ax + by + cz)}}$

From equation 1, 2 and 3 , we can write:-

$\\ \sf{\implies{(x + y + z)( ak + bk + ck)= k(ax + by + cz)}}$

$\\ \sf{\implies{k(x + y + z)(a + b + c) = k(ax + by + cz)}}$

Removing k from both sides:-

$\\ \sf{\implies{(x + y + z)(a + b + c) = (ax + by + cz)}}$

$\\ \sf{\therefore{\red{(x + y + z)(a + b + c) = (ax + by + cz)}}}$

Hence, showed!