Question

If $xy = p,xz = {p^2}$ and $yz = {p^3}$, also $x + y + z = 13$ and ${x^2} + {y^2} + {z^2} = 91$, then $\dfrac{z}{y}$ is equal to:
A. 3
B. $\dfrac{7}{3}$
C. 13
D. $\dfrac{{13}}{3}$

Answer 1

Hint: Substitute the given values in the formula ${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca$ and simplify it. It will give the polynomial of degree three. Use the hit and trial method to find the factors of p. The other two factors of p will not be real. So discard them. Now find the value of $\dfrac{z}{y}$ by multiplying numerator and denominator by x. Then, substitute the values and solve it to get the result.

Complete step-by-step solution:
Given: - $xy = p,xz = {p^2}$ and $yz = {p^3}$.
$x + y + z = 13$
${x^2} + {y^2} + {z^2} = 91$
As we know that ${\left( {a + b + c} \right)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca$.
Substitute $a = x,b = y$ and $c = z$ in the formula,
${\left( {x + y + z} \right)^2} = {x^2} + {y^2} + {z^2} + 2xy + 2yz + 2zx$
Now put the given values in the equation,
${13^2} = 91 + 2p + 2{p^3} + 2{p^2}$
Move 91 to the other side,
$2{p^3} + 2{p^2} + 2p = 169 - 91$
Take 2 commons from the left side and subtract 91 from 169 on the right side,
$2\left( {{p^3} + {p^2} + p} \right) = 78$
Divide both sides by 2,
${p^3} + {p^2} + p = 39$
Move 39 to the left side of the equation,
${p^3} + {p^2} + p - 39 = 0$
Use hit and trial method to find one root of the equation,
Put $p = 1$,
${1^3} + {1^2} + 1 - 39 = - 36 \ne 0$
So, 1 is not the root.
Put $p = 2$,
${2^3} + {2^2} + 2 - 39 = - 25 \ne 0$
So, 2 is not the root.
Put $p = 3$,
${3^3} + {3^2} + 3 - 39 = 0 = 0$
So, 3 is the root.
Divide the equation (1) by $\left( {p - 3} \right)$,
$
  \left. {p - 3} \right){p^3} + \,\,\,{p^2} + \,\,\,\,\,\,p - 39\left( {{p^2} + 4p + 13} \right. \\
  \,\,\,\,\,\,\,\,\,\,\,\,\underline {{p^3} - 3{p^2}} \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,4{p^2} + \,\,\,\,\,\,p \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline {4{p^2} - 12p} \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,13p - 39 \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\underline {13p - 39} \\
  \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,0 \\
$
So, the factors are,
$\left( {p - 3} \right)\left( {{p^2} + 4p + 13} \right) = 0$
Since 13 is a prime number, so it is not possible to break 4 in such a way that its multiple will be 13.
So, the real root of p is 3.
To find the value of $\dfrac{z}{y}$, multiply numerator and denominator by $x$.
$\dfrac{z}{y} = \dfrac{z}{y} \times \dfrac{x}{x} = \dfrac{{zx}}{{xy}}$
Now, substitute the value of $zx$ and $xy$,
$\dfrac{z}{y} = \dfrac{{{p^2}}}{p}$
Cancel out the common terms from numerator and denominator,
$\dfrac{z}{y} = p = 3$
Thus, the value of $\dfrac{z}{y}$ is 3.

Hence, option (A) is the correct answer.

Note: The algebraic equations which are valid for all values of variables in them are called algebraic identities. They are also used for the factorization of polynomials. In this way, algebraic identities are used in the computation of algebraic expressions and solving different polynomials.
Some Standard Algebraic Identities are:
Identity I: \[{{\left( a+b \right)}^{2}}={{a}^{2}}+2ab+{{b}^{2}}\]
Identity II: \[{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}\]
Identity III: \[{{a}^{2}}-{{b}^{2}}=\left( a+b \right)\left( ab \right)\]
Identity IV: \[\left( x+a \right)\left( x+b \right)={{x}^{2}}+\left( a+b \right)x+ab\]
Identity V: \[{{\left( a+b+c \right)}^{2}}={{a}^{2}}+{{b}^{2}}+{{c}^{2}}+2ab+2bc+2ca\]
Identity VI: \[{{\left( a+b \right)}^{3}}={{a}^{3}}+{{b}^{3}}+3ab\left( a+b \right)\]
Identity VII: \[{{\left( a-b \right)}^{3}}={{a}^{3}}-{{b}^{3}}-3ab\left( a-b \right)\]
Identity VIII: \[{{a}^{3}}+{{b}^{3}}+{{c}^{3}}-3abc=\left( a+b+c \right)({{a}^{2}}+{{b}^{2}}+{{c}^{2}}ab-bc-ca)\]