If $$(27x^3 - 343y^3) \div (3x - 7y) = Ax^2 + By^2 + 7Cyx$$, then the value of (4A - B + 5C)is:

As per the given question,

$$(27x^3 - 343y^3) \div (3x - 7y) = Ax^2 + By^2 + 7Cyx$$

We know that $$a^3-b^3=(a-b)(a^2+b^2+ab)$$

Now, $$(27x^3 - 343y^3) \div (3x - 7y) = (3x)^3-(7y)^3$$

$$=(3x-7y)(9x^2+49y^2+21xy)\div (3x-7y)=(9x^2+49y^2+21xy)$$

Hence $$9x^2+49y^2+7\times 3 xy=Ax^2 + By^2 + 7Cyx$$

By comparing of left and right coefficient of respective elements.

A=9, B=49, C=3

Now substituting the values $$(4A - B + 5C)=4\times 9-49+5\times 3=36+15-49=51-49=2$$