To solve this problem, we need to find the value of the expression:
\(ab(a + b) + bc(b + c) + ca(c + a) + 3abc\)
given the conditions:
- \(a + b + c = 13\)
- \(ab + bc + ca = 4\)
We can start by expanding the given expression:
\(ab(a + b) + bc(b + c) + ca(c + a) + 3abc = a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2 + 3abc\)
We can rewrite the expression as:
\(a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2 + 3abc = a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2 + abc + 2abc\)
Notice the terms can be rearranged into \(ab(a + b) + bc(b + c) + ca(c + a) = abc + (a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2 + 2abc)\).
By substituting \(a + b + c\) and \(ab + bc + ca\), we can solve:
\((a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)\)
We already know \(a + b + c = 13\) and \(ab + bc + ca = 4\). Using these values, calculate:
\(169 = a^2 + b^2 + c^2 + 8\)
Thus, \(a^2 + b^2 + c^2 = 161\).
Now rearrange \(a^2 + b^2 + c^2\) in terms and plug it back:
\(a^2b + ab^2 + b^2c + bc^2 + c^2a + ca^2\) forms part of \((ab + bc + ca)(a + b + c) - 3abc\).
Calculate \((ab + bc + ca)(a + b + c)\):
\(= (4)(13) = 52\)
Therefore, expression simplifies as:
\(ab(a + b) + bc(b + c) + ca(c + a) + 3abc = 52 - 3abc + 3abc = 52\)
Therefore, the value of the given expression is 52.
Important Questions from Algebra (Notes)
- What is the remainder when 2023²⁰²⁴ + 2025²⁰²⁴ is divided by 2024?
In an examination, a student scores 4 marks for every correct answer and loses 1 mark for every wrong answer. If she/he attempts all 60 questions and secures 130 marks, the number of questions she/he attempts wrongly, are?
Match List-I with List-II
List-1 List-II (A) If $\begin{bmatrix}\lambda-1 & 0 \\ 0 & \lambda-1 \end{bmatrix} $, then $\lambda$ is (I) 0 (B) If A=$ \begin{bmatrix}1 & 2 \\2 & 4 \end{bmatrix} $, then $\Delta$ is (II) 1 (C) If A = $ \begin{bmatrix}1 & 0 \\0 & \frac{1}{2} \end{bmatrix} $, then $|A^{-1}|$ is (III) -2 (D) If $ \begin{bmatrix}a+1 & 1 \\1 & 2 \end{bmatrix} = \begin{bmatrix}-1 & 1 \\1 & 2 \end{bmatrix} $, then a is (IV) 2 Choose the correct answer from the options given below:
- If (x - 1) is a factor of $2x^2 - 5x + k = 0$, then the value of k is:
- If $x = (2+\sqrt{3})^{\frac{1}{3}} + (2+\sqrt{3})^{-\frac{1}{3}}$ and $x^3-3x + k = 0$, then the value of k is:
