Question

j) (-3)^4 - (-2)^3^0 ÷ (-4)^3 - (-3)^2^0

Explanation:

Step1: Analyze the base of the first exponent

Calculate \((-3)^4\) and \((-2)^3\). \((-3)^4 = (-3)\times(-3)\times(-3)\times(-3)=81\), \((-2)^3 = (-2)\times(-2)\times(-2)= - 8\). Then \((-3)^4-(-2)^3=81 - (-8)=81 + 8 = 89\). So the first part \([(-3)^4-(-2)^3]^0 = 89^0\).

Step2: Analyze the base of the second exponent

Calculate \((-4)^3\) and \((-3)^2\). \((-4)^3=(-4)\times(-4)\times(-4)=-64\), \((-3)^2 = (-3)\times(-3)=9\). Then \((-4)^3-(-3)^2=-64 - 9=-73\). So the second part \([(-4)^3-(-3)^2]^0=(-73)^0\).

Step3: Apply the zero - exponent rule

Recall the zero - exponent rule: \(a^0 = 1\) for \(a
eq0\). Since \(89
eq0\) and \(-73
eq0\), we have \(89^0 = 1\) and \((-73)^0 = 1\).

Step4: Perform the division

Now we need to calculate \(1\div1\). According to the division rule of real numbers, any non - zero number divided by itself is \(1\). So \(1\div1 = 1\).

Answer:

\(1\)