Write the following in the expanded form \[{(3x + 7y + 5z)^2}\]

Hint: In this we can use the various algebraic formulas as \[{(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca\] and hence on expanding the above given value, we can get our required answer.

Complete step-by-step answer:
As the given term for the expansion is \[{(3x + 7y + 5z)^2}\]
Now, using the formula for the expansion as \[{(a + b + c)^2} = {a^2} + {b^2} + {c^2} + 2ab + 2bc + 2ca\]
Hence, on expanding we can see that
 \[ \Rightarrow {(3x + 7y + 5z)^2} = {(3x)^2} + {(7y)^2} + {(5z)^2} + 2(3x)(7y) + 2(7y)(5z) + 2(5z)(3z)\]
Now on expanding the terms of bracket inside R.H.S, we get,
 \[ \Rightarrow {(3x + 7y + 5z)^2} = 9{x^2} + 49{y^2} + 25{z^2} + 42xy + 70yz + 30zx\]
Hence, above is our required expanded form for the given term.
Additional information:
Some other important expansions are,
For squaring the terms of binomials,
 \[{(a \pm b)^2} = {a^2} \pm 2ab + {b^2}\]
For cubing the terms of binomial,
 \[{(a \pm b)^3} = {a^3} \pm 3ab(a + b) + {b^3}\]

Note: Algebra is the study of mathematical symbols and the rules for manipulating these symbols. It is a unifying thread of almost all of mathematics. It includes everything from elementary equation solving to the study of abstractions such as groups, rings, and fields.
The way we use the shortcut is to follow three simple steps.
Step \[1\]: Square the first term of the binomial.
 Step \[2\]: Multiply the first term and last term of the binomial together and then double that quantity (in other words multiply by \[2\]).
Step \[3\]: Square the last term of the binomial.
If you forget the formula for expansion for 3 variables then you can simply multiply them twice as shown below,
 \[{(3x + 7y + 5z)^2}\]
This can be written as
 \[ \Rightarrow (3x + 7y + 5z) \times (3x + 7y + 5z)\]
On multiplying we get,
 \[ \Rightarrow 3x(3x) + 3x(7y) + 3x(5z) + 7y(3x) + 7y(7y) + 7y(5z) + 5z(3x) + 5z(7y) + 5z(5z)\]
On simplification we get,
 \[ \Rightarrow 9{x^2} + 21xy + 15xz + 21xy + 49{y^2} + 35yz + 15xz + 35yz + 25{z^2}\]
On adding like terms we get,
 \[ \Rightarrow 9{x^2} + 49{y^2} + 25{z^2} + 42xy + 30xz + 70yz\]
Which is the same as the answer obtained.
So solving either way will yield the same answer, and hence we can use any method to solve, it is just that the first method takes less number of steps and is easy for solution.