How do you find all the solutions to $ {x^4} - 81 = 0 $ ?

Hint: In this question, we are given a polynomial equation, the highest exponent in the given equation is 4, so the degree of the equation is 4, and hence the given equation is quadratic and has four solutions. In this question, we will simply bring 81 to the other side of the equal sign and divide both sides then by prime factorization of 81; we will write as a square of some number, then square rooting both sides of the equation we get two numbers. Again expressing the obtained numbers as a square of some other numbers, we will get four values of x. Thus, the values obtained will be the roots of the given equation.

Complete step-by-step answer:
We are given –
 $
  {x^4} - 81 = 0 \\
   \Rightarrow {x^4} = 81 \\
   \Rightarrow {({x^2})^2} = {( \pm 9)^2} \\
   \Rightarrow {x^2} = \pm 9 \;
  $
Now, $ {x^2} = 9 $ and $ {x^2} = - 9 $
 $
   \Rightarrow {x^2} = {( \pm 3)^2},\,{x^2} = {( \pm 3\sqrt { - 1} )^2} \\
   \Rightarrow x = \pm 3,\,x = \pm 3i \;
  $
Hence the factors of the equation $ {x^4} - 81 $ are $ x - 3 = 0,\,\,x + 3 = 0,\,\,x - 3i = 0\,\,and\,\,x + 3i = 0 $ .
So, the correct answer is “ $ x - 3 = 0,\,\,x + 3 = 0,\,\,x - 3i = 0\,\,and\,\,x + 3i = 0 $ ”.

Note: By the solutions of the equation, we mean the values of the x for which the function given as $ f(x) = {x^4} - 81 $ has a value zero or when we plot this function on the graph, we see the points on which the y-coordinate is zero are the solutions of this equation, thus they are simply the x-intercepts. We can also solve the given equation by using an arithmetic identity, according to which the difference of the square of one number a and the square of another number b is equal to the product of the difference of a and b and the sum of a and b, that is, $ {a^2} - {b^2} = (a - b)(a + b) $ .