Answer:
Proved below
Explanation:
We want to prove that;
√(2yx^(-1)) × √(4xz^(-1)) × √(8zy^(-1)) = 8
Now, the √ symbol is also written by denoting the numbers attached to it as resisted to the power of ½. Thus, we have;
[(2yx^(-1))^(½)] × [(4xz^(-1))^(½)] × [(8zy^(-1))^(½)]
Simplifying this gives;
2^(½) × y^(½) × x^(-½) × 4^(½) × x^(½) × z^(-½) × 8^(½) × z^(½) × y^(-½)
Simplifying further, let's write 8 and 4 in terms of 2. Thus;
2^(½) × y^(½) × x^(-½) × (2²)^(½) × x^(½) × z^(-½) × (2³)^(½) × z^(½) × y^(-½)
This gives;
2^(½) × y^(½) × x^(-½) × (2¹) × x^(½) × z^(-½) × (2^(3/2)) × z^(½) × y^(-½)
Collecting like terms and applying laws of indices gives;
2^(½ + 1 + (3/2)) × y^(½ - ½) × x^(-½ + ½) × z^(-½ + ½)
This gives;
8y^(0) × x^(0) × y^(0)
Any Number raised to the power of zero equals 1. Thus;
8y^(0) × x^(0) × y^(0) = 8