Question

If the expression \[x+y+z=0\], then the value of \[\dfrac{{{x}^{2}}}{yz}+\dfrac{{{y}^{2}}}{zx}+\dfrac{{{z}^{2}}}{xy}\] is
(a) \[2\sqrt{3}\]
(b) \[\sqrt{3}\]
(c) 3
(d) \[4\sqrt{3}\]

Answer 1

Hint: Simplify the given equation by taking LCM of terms and then use the identity that if \[x+y+z=0\], then we have \[{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=0\] to find the value of the given expression.

Complete step-by-step answer:
We know that \[x+y+z=0\]. We have to calculate the value of \[\dfrac{{{x}^{2}}}{yz}+\dfrac{{{y}^{2}}}{zx}+\dfrac{{{z}^{2}}}{xy}\].
We will simplify the terms of the given equation by taking LCM of the given equation.
Thus, by taking LCM, we have \[\dfrac{{{x}^{2}}}{yz}+\dfrac{{{y}^{2}}}{zx}+\dfrac{{{z}^{2}}}{xy}=\dfrac{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}}{xyz}\]
We know the identity that if \[x+y+z=0\], then we have \[{{x}^{3}}+{{y}^{3}}+{{z}^{3}}-3xyz=0\].
We can rewrite the above equation as \[{{x}^{3}}+{{y}^{3}}+{{z}^{3}}=3xyz\].
Substituting the above equation in the expression \[\dfrac{{{x}^{2}}}{yz}+\dfrac{{{y}^{2}}}{zx}+\dfrac{{{z}^{2}}}{xy}=\dfrac{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}}{xyz}\], we have \[\dfrac{{{x}^{2}}}{yz}+\dfrac{{{y}^{2}}}{zx}+\dfrac{{{z}^{2}}}{xy}=\dfrac{{{x}^{3}}+{{y}^{3}}+{{z}^{3}}}{xyz}=\dfrac{3xyz}{xyz}\].
Thus, we have \[\dfrac{{{x}^{2}}}{yz}+\dfrac{{{y}^{2}}}{zx}+\dfrac{{{z}^{2}}}{xy}=3\].
Hence, the value of \[\dfrac{{{x}^{2}}}{yz}+\dfrac{{{y}^{2}}}{zx}+\dfrac{{{z}^{2}}}{xy}\] is 3 when we have \[x+y+z=0\], which is option (c).

An algebraic identity is an equality that holds for all possible values of its variables. We can prove each of the identities by performing basic algebraic operations such as addition, multiplication, subtraction and division. They are used for the factorization of the polynomials. That’s why they are useful in the computation of algebraic expressions. An algebraic expression differs from an algebraic identity in the way that the value of an algebraic expression changes with the change in variables. However, an algebraic identity is an equality which holds for all possible values of variables.


Note: We can prove the given identity by taking the cube of the equation \[x+y+z=0\] on both sides and simplifying it using the identity \[{{\left( a+b+c \right)}^{3}}={{a}^{3}}+{{b}^{3}}+{{c}^{3}}+3{{a}^{2}}b+3{{a}^{2}}c+3{{b}^{2}}c+3{{b}^{2}}a+3{{c}^{2}}a+3{{c}^{2}}b+6abc\]. All these identities used in this question are algebraic identities. We can also solve this question using the identity \[{{a}^{3}}+{{b}^{3}}+{{x}^{3}}={{\left( a+b+c \right)}^{2}}-3ab\left( a+b \right)-3bc\left( b+c \right)-3ac\left( a+c \right)-6abc\] to simplify the terms and calculate the value of the given expression.