Help simplifying this algebraic expression please

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The discussion revolves around simplifying algebraic expressions. In the first case, the user initially miscalculated the expression 3x^3 + 3/8x^3 but later clarified that the correct simplification yields 3/2 x^3. For the second expression, the user struggled with the operation involving 9x^3/2 - 1/x^3/2, realizing that they cannot simply square individual terms. They also sought clarification on whether -3x^0 equals -3 or 1, confirming that it simplifies to -3. Overall, the conversation emphasizes the importance of proper algebraic manipulation and understanding of exponent rules.
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14) here i tried 3x^3 + 3/8x^3 to to get 27/8x^3 but the answer is 3/2 x^3, why?
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15) here i did 9x^3/2 - 1/x^3/2 to get 9x^9/4 - 1/x^3/2 but that's not in the right form, how do i do it correctly?
 
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homeworkhelpls said:
View attachment 31519414) here i tried 3x^3 + 3/8x^3 to to get 27/8x^3 but the answer is 3/2 x^3, why?
Did you take the cube root of the constant multiplier?
homeworkhelpls said:
View attachment 31519615) here i did 9x^3/2 - 1/x^3/2 to get 9x^9/4 - 1/x^3/2 but that's not in the right form, how do i do it correctly?
That is not correct. You can not simply square the individual terms. Do you know what the general result of ##(a-b)^2## is? Multiply out the square to see what is right.
 
ok then for 15 i expanded the brackets and got 9x^3/2 -3x^0 - 3x^0 + x^-3/2, I am confused if -3x^0 is equal to -3 or 1, please explain
 
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FactChecker said:
Did you take the cube root of the constant multiplier?

That is not correct. You can not simply square the individual terms. Do you know what the general result of ##(a-b)^2## is? Multiply out the square to see what is right.
and for 14 i got it now after doing 3 x by cubed root of 3 x by x^3 all over 2 to get 3/2 x^3, thanks :)
 
homeworkhelpls said:
ok then for 15 i expanded the brackets and got 9x^3/2 -3x^0 - 3x^0 + x^-3/2, I am confused if -3x^0 is equal to -3 or 1, please explain
A product of both. ##-3x^0 = -3 \cdot x^0 = -3 \cdot 1 = -3##
 
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Thread 'Find the polar form of a complex number'

The working out suggests first equating ## \sqrt{i} = x + iy ## and suggests that squaring and equating real and imaginary parts of both sides results in ## \sqrt{i} = \pm (1+i)/ \sqrt{2} ## Squaring both sides results in: $$ i = (x + iy)^2 $$ $$ i = x^2 + 2ixy -y^2 $$ equating real parts gives $$ x^2 - y^2 = 0 $$ $$ (x+y)(x-y) = 0 $$ $$ x = \pm y $$ equating imaginary parts gives: $$ i = 2ixy $$ $$ 2xy = 1 $$ I'm not really sure how to proceed from here.
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