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Can you explain how expression follows the grammar of the lambda calculus?

$$\lambda x.x((\lambda y.yy)x)x = λx.x(xx)x$$

I am not sure why we have the parentheses following the $.x$ (on both sides of the equation).

Can you explain how this expression follows the grammar of the lambda calculus?

$$\lambda x.x((\lambda y.yy)x)x = λx.x(xx)x$$

I am not sure why we have the parentheses following the $.x$ (on both sides of the equation).

Can you explain how does this expression follow the grammar of lambda calculus?

$\lambda x.x((\lambda y.yy)x)x = λx.x(xx)x$

I am not sure why we have the parentheses following the .x (on both sides of the equation).

Can you explain how expression follows the grammar of the lambda calculus?

$$\lambda x.x((\lambda y.yy)x)x = λx.x(xx)x$$

I am not sure why we have the parentheses following the $.x$ (on both sides of the equation).

Can you explain how does this expression follow the grammar of lambda calculus?

λx.x((λy.yy)x)x = λx.x(xx)x

I am not sure why we have the parentheses following the .x (on both sides of the equation).

Can you explain how does this expression follow the grammar of lambda calculus?

$\lambda x.x((\lambda y.yy)x)x = λx.x(xx)x$

I am not sure why we have the parentheses following the .x (on both sides of the equation).