The number of triples (x, y, z) of real numbe | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given the equation $x^4+y^4+z^4+1=4xyz$.By the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality for four positive numbers $x^4, y^4, z^4,

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(B) Given the equation $x^4+y^4+z^4+1=4xyz$.By the Arithmetic Mean-Geometric Mean $(AM \geq GM)$ inequality for four positive numbers $x^4, y^4, z^4, 1$:$\frac{x^4+y^4+z^4+1}{4} \geq \sqrt[4]{x^4 y^4 z^4 \cdot 1}$Substituting the given equation $x^4+y^4+z^4+1 = 4xyz$:$\frac{4xyz}{4} \geq |xyz|$$xyz \geq |xyz|$This inequality holds if and only if $xyz \geq 0$. For equality to hold in $AM \geq GM$,all terms must be equal:$x^4 = y^4 = z^4 = 1$This implies $|x| = 1, |y| = 1, |z| = 1$. Thus,$x, y, z \in \{1, -1\}$.Substituting these into the original equation $x^4+y^4+z^4+1=4xyz$:$1+1+1+1 = 4xyz$ $\Rightarrow 4 = 4xyz$ $\Rightarrow xyz = 1$.The possible triples $(x, y, z)$ such that their product is $1$ are:$(1, 1, 1), (1, -1, -1), (-1, 1, -1), (-1, -1, 1)$.There are $4$ such triples.

The number of triples $(x, y, z)$ of real numbers satisfying the equation $x^4+y^4+z^4+1=4xyz$ is

Similar Questions

If $a, b, c$ are distinct positive real numbers and $a^2+b^2+c^2=1$,then the value of $ab+bc+ca$ is

The equation $\left(x^4+1\right)=\frac{1}{a}(x+1)^4$ is a reciprocal equation:

Let $p, q$ be real numbers. If $\alpha$ is a root of $x^{2}+3 p^{2} x+5 q^{2}=0$,$\beta$ is a root of $x^{2}+9 p^{2} x+15 q^{2}=0$ and $0 < \alpha < \beta$,then the equation $x^{2}+6 p^{2} x+10 q^{2}=0$ has a root $\gamma$ that always satisfies:

The minimum value of $f(x) = \frac{x^2-2x+3}{x^2-4x+7}$ is

If $\alpha, \beta$ are the roots of $x^2-a(x-1)+b=0$,then the value of $\frac{1}{\alpha^2-a \alpha}+\frac{1}{\beta^2-a \beta}+\frac{2}{a+b}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module