Question

what is the value of $(x^{0}y^{2}z^{3})^{0}$?
options: 0, $y^{2}z^{3}$, 1, $x^{0}y^{0}z^{0}$

Explanation:

Step1: Recall the zero - exponent rule

The zero - exponent rule states that for any non - zero real number \(a\), \(a^{0}=1\). Also, when we have a power of a product \((abc)^{n}=a^{n}b^{n}c^{n}\), and here we have \((x^{0}y^{2}z^{3})^{0}\).

Step2: Apply the power of a product rule

Using the power of a product rule \((a^{m}b^{n}c^{p})^{q}=a^{mq}b^{nq}c^{pq}\), for \((x^{0}y^{2}z^{3})^{0}\), we get \(x^{0\times0}y^{2\times0}z^{3\times0}\).

Step3: Simplify the exponents

Simplifying the exponents: \(0\times0 = 0\), \(2\times0=0\), \(3\times0 = 0\). So we have \(x^{0}y^{0}z^{0}\).

Step4: Apply the zero - exponent rule again

Since for any non - zero \(a\), \(a^{0}=1\) (and assuming \(x\), \(y\), \(z\) are non - zero, if any of them is zero, the original expression \((x^{0}y^{2}z^{3})^{0}\) is still defined as 1 because \(0^{0}\) in the context of exponentiation rules for such cases (where we are raising a product to the zero power) is considered to be 1), \(x^{0}=1\), \(y^{0}=1\), \(z^{0}=1\). Then \(x^{0}y^{0}z^{0}=1\times1\times1 = 1\).

Answer:

C. 1