Question

1. If x+y+z=0, then show that x^2+y^2+z^2=-2(xy+yz+zx).
7

2. If a=3+b, prove that a^3-b^3-9ab-27.

3. Factorise 9 x^2-12x+4.

4. Factorise 343+27t^3

Please give the answer​

Answer 1

Answer:

Step-by-step explanation:

Answer 2

1. If x+y+z=0, then show that x²+y²+z²= -2(xy+yz+zx).

➡️➡️

Here,

$\rm \huge{}\to x+y+z=0 \\ \\ \boxed{ \bf squaring \: \: both \: \: sides} \\ \\ \Large \tt \to{(x + y + z )}^{2} = {0}^{2}$

$\small{ \rm{ \blue{ Identity \: Required :} \green \to }} \rm{ \orange{{(a + b+ c)}^{2} }= \purple{ {a}^{2} + {b}^{2} + {c}^{2} + 2(ab + bc + ac)} }$

So,

$\to \rm \small {x}^{2} + {y}^{2} + {z}^{2} + 2(xy + yz + xz) = 0 \\ \to \rm \small {x}^{2} + {y}^{2} + {z}^{2} = 0 - \{ 2(xy + yz + xz) \} \\ \to\underline{ \underline{ \rm \small {x}^{2} + {y}^{2} + {z}^{2} = \: \: - 2(xy + yz + xz) }}$

$\tt \orange{ HENCE<strong> </strong>\: \: \: PROVED }\red{!!}$

2. If a=3+b, Prove that a³ -b³ - 9ab = 27.

➡➡

Here,

$\rm \huge {a=3+b} \\ \\ \boxed{ \text{ Cubing on both sides, }} \\ \tt {a}^{3} = {(3 + b)}^{3} \\ \small\tt {a}^{3} = {(3)}^{3} + {(b)}^{3} + 3(3)(b) \{3 + b \} \\ \tt {a}^{3} = 27 + {b}^{3} + 9b(3 + b) \\ \tt {a}^{3} = 27 + {b}^{3} + 9b(a) \: \: \: \: \: \: \: \: \{ \because \: a = 3 + b\} \\ \rm {a}^{3} = 27 + {b}^{3} + 9ab \\ \boxed{ \rm transposing \: \: terms} \\ \rm \underline{ \underline{{a}^{3} - {b}^{3} - 9ab = 27}}$

$\tt \green{ HENCE \: \: \: PROVED} \pink{ !! }$

3. Factorise 9x² - 12x + 4.

$\to \rm {9x}^{2} - 12x + 4 \\ \to \rm {9x}^{2} - 6x - 6x + 4 \\ \to \rm3x(3x - 2) - 2(3x - 2) \\ \to \underline{\underline{ \rm \red{(3x - 2)(3x - 2)}}}$

4. Factorise 343 + 27t³

Here,

• Polynomial is in form of a³ + b³ , the identity used will be

$\tt \orange{{a}^{3} + {b}^{3} }= \purple{(a + b)( {a}^{2} + {b}^{2} - ab )}$

Now,

$\to\rm343 + {27t}^{3} \\ \to \rm{(7)}^{3} + {(3t)}^{3} \\ \to \rm(7 + 3t) \{ {(7)}^{2} + {(3t)}^{2} - (7)(3t) \} \\ \to \rm\underline{\underline{ \red{(7 +3 t)(49 + {9t}^{2} - 21t)}}}$