The cube of a certain binomial is $8y^3-36y^2+54y-27$. Find the binomial.
I know that $(a + b)^3 = a^3 + 3a^2b + 3ab^2 + b^3$ and that$(a - b)^3 = a^3 - 3a^2b + 3ab^2 - b^3$ but don't know how to go further...
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1$\begingroup$ If you have $a^3=8, b^3=(-)27$ in $a^3x^3+\dots+(-)27$, what do you think could be the solution? $\endgroup$– abiessuCommented Sep 21, 2016 at 3:09
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4 Answers
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Assume that the degree of $a$ is larger than the degree of $b$ so that its expansion is ordered in descending powers of $y$. Then we can match the highest and lowest degree terms: $$ a^3 = 8y^3 \implies a = 2y \\ b^3 = -27 \implies b = -3 $$ Finally, verify that our guess is correct by expanding $(2y - 3)^3$ and checking that it matches the original polynomial.
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Think about what the cube root would look like and how you can algebraically represent its cube. What kind of binomial, when cubed, would yield a cubic in one variable?
answered Sep 21, 2016 at 3:04
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$\begingroup$ @thunderbolt This better? $\endgroup$ Commented Sep 21, 2016 at 3:10
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$\begingroup$ binomial with second term 0 degree? $\endgroup$ Commented Sep 21, 2016 at 3:13
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$\begingroup$ @thunderbolt Yes. Now think about what of linear binomial, when cubed, is $8y^3-36y^2+54y-27$. $\endgroup$ Commented Sep 21, 2016 at 3:14
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Simply look at the last two terms. $$(a-b)^3=a^3-3a^2b+3ab^2-b^3\tag{1}$$ and from the binomial $$8y^3-36y^2+54y-27\tag{2}$$
We see that $$\begin{cases}8y^3=a^3\\27=b^3\end{cases}\tag{3}$$
Solving, we see that $a=2y$ and $b=3$. So the binomial factors into $$(2y-3)^3\tag{4}$$
Expanding out $(4)$ to check, we get: $$8y^3-27-3(2y)^2(3)+3(2y)(9)\tag{5}\\=8y^3-9\cdot4y^2+54y-27\\=8y^3-36y^2+54y-27$$
Which is equal to $(2)$.
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We can see that the plus and minus pattern follows the $(a-b)^3$ expansion. Then simply equate coefficients in any two terms:
$8y^3 = a^3 \implies a = 2y$
$27 = b^3 \implies b = 3$
So the cube root is $(2y-3)$
