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I have the determinant :

\begin{vmatrix} 1 &1 &1 \\ a &b &c \\ a^3 &b^3 &c^3 \\ \end{vmatrix}

How do I prove that this determinant is equal to

$$ (a-b)(b-c)(c-a)(a+b+c) $$

Rodrigo de Azevedo
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Bogdan Pop
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5 Answers5

6

HINT

In questions like these, our main goal is to create 2 zeros in a row or column, if possible, along which we can expand the determinant.

Rodrigo de Azevedo
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Gathdi
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Expand this

$$\begin{vmatrix} 1 &1 &1 \\ a &b &c \\ a^3 &b^3 &c^3 \\ \end{vmatrix}$$

to an expanded polynomial expression. Now expand this $$ (a-b)(b-c)(c-a)(a+b+c) $$ to an expanded polynomial expression.

Now check that they're both equal.

Anon
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  • What I did: the determinant is equal to $$ \begin{vmatrix} b &c \ b^3 &c^3 \ \end{vmatrix} - \begin{vmatrix} a &c \ a^3 &c^3 \ \end{vmatrix} + \begin{vmatrix} a &b \ a^3 &b^3 \ \end{vmatrix} $$ – Bogdan Pop Jun 11 '16 at 12:41
  • Then I got to $$ (bc^3-b^3c)-(ac^3-a^3c)+(ab^3-a^3b) $$ – Bogdan Pop Jun 11 '16 at 12:41
  • How do I factor this to the given formula... that's where I remained stucked – Bogdan Pop Jun 11 '16 at 12:42
  • You can work in both directions. Instead of factoring that there, just expand the other expression. – Anon Jun 11 '16 at 12:43
  • Expanding that formula takes some time, isn't there a more shorter way of doing it? – Bogdan Pop Jun 11 '16 at 12:47
  • @BogdanPop Look at the expression you got. If you put $a=b$ it is obviously 0, so it must have a factor $a-b$. Notice that you could also deduce that the det must have a factor $(a-b)(b-c)(c-a)$ just by looking at it. – almagest Jun 11 '16 at 12:47
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HINT:

Use $$C_1'=C_1-C_3$$ and $$C_2'=C_2-C_3$$

where $C_r$ is the $r$th column, $C_r'$ is the resultant $r$th column

See also: Factorise the determinant $\det\Bigl(\begin{smallmatrix} a^3+a^2 & a & 1 \\ b^3+b^2 & b & 1 \\ c^3+c^2 & c &1\end{smallmatrix}\Bigr)$

Community
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lab bhattacharjee
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I'll start from where you ended in the comments:

$$bc^3-cb^3-(ac^3-ca^3)+ab^3-ba^3$$ $$bc^3-cb^3-ac^3+ca^3+ab^3-ba^3$$

Now, we need to factor this polynomial. We can do this by trying to group certain terms together and find patterns.

  • There are two $a^3$ terms: $ca^3-ba^3$. This can be factored into $a^3(c-b)=-a^3(b-c)$
  • There are two $a$ terms: $ab^3-ac^3$. This can be factored into $a(b^3-c^3)=a(b-c)(b^2+cb+c^2)$
  • There are two terms without an $a$: $bc^3-cb^3$. This can be factored into $bc(c^2-b^2)=bc(c+b)(c-b)=-cb(b+c)(b-c)$

Thus, we can factor out a $b-c$ from all of the terms, giving us:

$$(b-c)(-a^3+a(b^2+cb+c^2)-cb(b-c))=(b-c)(ab^2+abc+ac^2-cb^2-c^2b-a^3)$$

Now, let's try to do what we did before, except now with $b$ instead of $a$.

  • There are two $b^2$ terms: $ab^2-cb^2$. This can be factored into $b^2(a-c)=-b^2(c-a)$.
  • There are two $b$ terms: $abc-c^2b$. This can be factored into $bc(a-c)=-bc(c-a)$.
  • There are two terms without a $b$: $ac^2-a^3$. This can be factored into $a(c^2-a^2)=a(c+a)(c-a)$.

Thus, we can factor out a $c-a$, giving us:

$$(b-c)(c-a)(-b^2-bc+a(c+a))=(b-c)(c-a)(ac+a^2-b^2-bc)$$

Now, we can do the same thing we did before, but with $c$:

  • There are two $c$ terms: $ac-bc$. This can be factored into $c(a-b)$.
  • There are two terms without a $c$: $a^2-b^2$. This can be factored into $(a+b)(a-b)$.

Thus, we can factor out an $a-b$, giving us:

$$(a-b)(b-c)(c-a)(c+a+b)=(a-b)(b-c)(c-a)(a+b+c)$$

Noble Mushtak
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0

Hint: $a^3 -b^3$ $=$ $(a-b)(a^2 + +ab + b^2)$

Subtract column $1$ from column $2$ and then column $2$ from column $3$.

JMP
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Aakash Kumar
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